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Why do noble gases have positive electron gain enthalpies while halogens have highly negative values? | Noble gases possess a completely filled valence shell with a stable electron configuration of $$ns^2np^6$$. Adding an electron to a noble gas requires placing it in the next higher energy level (n+1 shell), which is energetically unfavorable and requires energy input. This results in a positive electron gain enthalpy. ... | Chemistry 11th |
What is electronegativity? How does it differ from electron gain enthalpy? | Electronegativity is a qualitative, relative property. It refers to the ability of an atom in a chemical bond to attract shared electrons. It has no units. Electron gain enthalpy is a quantitative, absolute property. It is the energy change that occurs when an isolated gaseous atom gains an electron. Its units are kJ/m... | Chemistry 11th |
What is valence? How is it related to the number of valence electrons in representative elements? | Valence is the combining power of an element. It determines the number of chemical bonds an atom can form. Valence electrons are the electrons in the outermost energy shell of an atom. For metals in groups 1, 2, and 13, the valence is equal to the number of valence electrons. For example, Sodium (Group 1) has 1 valence... | Chemistry 11th |
Distinguish between valence and oxidation state. Are they always the same? | Valence is the number of bonds an atom can form. It is a pure number and has no positive or negative sign. For example, the valence of carbon in ethane ($$C_2H_6$$) is 4, as it forms four bonds. The oxidation state (or oxidation number) represents the hypothetical charge an atom would have if all bonds were completely ... | Chemistry 11th |
Show by chemical equations that $\mathrm{Na_2O}$ is a basic oxide while $\mathrm{Cl_2O_7}$ is an acidic oxide. | A basic oxide is a metal oxide that reacts with water to form a base. Sodium oxide, an oxide of a metal, reacts with water to produce sodium hydroxide, which is a strong base. The chemical equation for the reaction of sodium oxide with water is: $$\mathrm{Na_2O(s) + H_2O(l) \rightarrow 2NaOH(aq)}$$ An acidic oxide is a... | Chemistry 11th |
How do shielding effect and nuclear charge together determine the effective nuclear charge experienced by valence electrons? | The total positive charge of the nucleus, determined by the number of protons, is the nuclear charge ($$Z$$). Inner-shell electrons, or core electrons, repel the outer-shell valence electrons. This repulsion is called the shielding effect ($$S$$). The valence electrons, therefore, experience a reduced attraction from t... | Chemistry 11th |
What is the diagonal relationship in the periodic table? Give an example of a pair of elements showing diagonal relationship. | The diagonal relationship describes the similarity in chemical properties between pairs of diagonally adjacent elements in the second and third periods. This relationship arises because these element pairs have a similar charge-to-radius ratio ($$\frac{q}{r}$$), which leads to similar polarizing power. Moving across a ... | Chemistry 11th |
The diameter of zinc atom is 2.6 Å. Calculate\
(a) radius of zinc atom in pm and\
(b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise. | The given diameter of the zinc atom is 2.6 Angstrom and the total length is 1.6 cm. The formula for the radius is $$r = d/2$$. The formula for the number of atoms is $$N = \frac{\text{Total Length}}{\text{Diameter}}$$. (a) First, convert diameter to picometers: $$2.6 \text{ Angstrom} \times \frac{100 \text{ pm}}{1 \t... | Chemistry 11th |
The threshold frequency $v_{0}$ for a metal is $7.0 \times 10^{14} \mathrm{~s}^{-1}$. Calculate the kinetic energy of an electron emitted when radiation of frequency $v=1.0 \times 10^{15} \mathrm{~s}^{-1}$ hits the metal. | The given data are the threshold frequency, $$v_0 = 7.0 \times 10^{14} \mathrm{~s}^{-1}$$, and the incident radiation frequency, $$v = 1.0 \times 10^{15} \mathrm{~s}^{-1}$$. Planck's constant is $$h = 6.626 \times 10^{-34} \mathrm{~J \cdot s}$$. The formula for the kinetic energy ($$K.E.$$) of a photoelectron is given ... | Chemistry 11th |
A microscope using suitable photons is employed to locate an electron in an atom within a distance of $0.1 \AA$. What is the uncertainty involved in the measurement of its velocity? | Given: Uncertainty in position, $$\Delta x = 0.1 \AA = 0.1 \times 10^{-10}$$ m. Mass of electron, $$m = 9.11 \times 10^{-31}$$ kg. Planck's constant, $$h = 6.626 \times 10^{-34}$$ J s. The Heisenberg Uncertainty Principle is $$\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$$. Since momentum $$p = mv$$, the uncertainty in m... | Chemistry 11th |
Calculate the wavelength, frequency and wavenumber of a light wave whose period is $2.0 \times 10^{-10} \mathrm{~s}$. | Given the period $$T = 2.0 \times 10^{-10} \text{ s}$$ and the speed of light $$c = 3.0 \times 10^8 \text{ m/s}$$. The formulas for frequency ($$\nu$$), wavelength ($$\lambda$$), and wavenumber ($$\nuÌ$$) are: $$\nu = \frac{1}{T}$$, $$\lambda = \frac{c}{\nu}$$, and $$\nuÌ = \frac{1}{\lambda}$$. First, calculate the f... | Chemistry 11th |
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with $n=4$ to an energy level with $n=2$ ? | The initial energy level is $$n_i = 4$$ and the final energy level is $$n_f = 2$$. The Rydberg constant is $$R_H = 1.097 \times 10^7 \text{ m}^{-1}$$. The Rydberg formula is used: $$\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$$. Substituting the values gives: $$\frac{1}{\lambda} = (1.097 \t... | Chemistry 11th |
What is the maximum number of emission lines when the excited electron of a H atom in $n=6$ drops to the ground state? | The initial energy level is $$n_2 = 6$$ and the final ground state energy level is $$n_1 = 1$$. The formula for the maximum number of emission lines is $$N = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$$. Substituting the values: $$N = \frac{(6 - 1)(6 - 1 + 1)}{2} = \frac{(5)(6)}{2} = \frac{30}{2} = 15$$. The maximum number o... | Chemistry 11th |
An atom of an element contains 29 electrons and 35 neutrons. Deduce\
(i) the number of protons and\
(ii) the electronic configuration of the element. | Given data: Number of electrons = 29, Number of neutrons = 35. For a neutral atom, the number of protons is equal to the number of electrons. Therefore, the number of protons is 29. The atomic number ($$Z$$) is 29, which identifies the element as Copper ($$\text{Cu}$$). Electrons fill orbitals in order of increasing en... | Chemistry 11th |
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit. | According to Bohr's second postulate, the angular momentum of an electron is quantized: $$L = mvr = \frac{nh}{2\pi}$$ where $$n$$ is the principal quantum number. According to the de Broglie hypothesis, the wavelength associated with the electron is given by: $$\lambda = \frac{h}{p} = \frac{h}{mv}$$ Rearranging the Boh... | Chemistry 11th |
Symbols ${ }_{35}^{79} \mathrm{Br}$ and ${ }^{79} \mathrm{Br}$ can be written, whereas symbols ${ }_{79}^{35} \mathrm{Br}$ and ${ }^{35} \mathrm{Br}$ are not acceptable. Answer briefly. | An element's symbol is defined by its atomic number ($$Z$$), which is the number of protons. The mass number ($$A$$) is the total number of protons and neutrons. For Bromine ($$\mathrm{Br}$$), the atomic number is always 35. Therefore, the subscript must be 35. The notation ${ }_{35}^{79} \mathrm{Br}$ is correct. Since... | Chemistry 11th |
Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?\
(i) $2 s$ and $3 s$,\
(ii) $4 d$ and $4 f$,\
(iii) $3 d$ and $3 p$. | For the 2s and 3s orbitals, the 2s orbital is closer to the nucleus. Electrons closer to the nucleus are less shielded and experience a greater effective nuclear charge. Therefore, 2s has a larger $$Z_{eff}$$. For the 4d and 4f orbitals, the principal quantum number ($$n=4$$) is the same. The effective nuclear charge i... | Chemistry 11th |
What are Dobereiner's Triads? Give two examples of triads with their atomic weights. | Dobereiner's Triads are groups of three elements that share similar chemical properties. The key principle is that the atomic weight of the middle element is approximately the arithmetic mean of the atomic weights of the other two elements. Example 1: The triad of Lithium ($$Li$$), Sodium ($$Na$$), and Potassium ($$K$$... | Chemistry 11th |
State Mendeleev's Periodic Law. On what basis did Mendeleev arrange the elements in his periodic table? | Mendeleev's Periodic Law states that the physical and chemical properties of the elements are a periodic function of their atomic masses. The primary basis for arranging the elements in his periodic table was the increasing order of their atomic mass. Mendeleev also grouped elements with similar chemical and physical p... | Chemistry 11th |
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination\
(s) has/have the same energy lists:
\begin{enumerate}
\item $n=4, l=2, m_{l}=-2, m_{\mathrm{s}}=-1 / 2$
\item $n=3, l=2, m_{l}=1, m_{\mathrm{s}}=+1 / 2$
\item $n=4, l=1, m_{l}=0, m_{s}... | The energy of an electron is determined by the principal quantum number, $$n$$, and the azimuthal quantum number, $$l$$. We calculate the value of ($$n+l$$) for each electron.
1. $$n=4, l=2 \implies n+l = 6$$
2. $$n=3, l=2 \implies n+l = 5$$
3. $$n=4, l=1 \implies n+l = 5$$
4. $$n=3, l=2 \implies n+l = 5$$
5. $$n=3, l=... | Chemistry 11th |
Why does the first period contain only 2 elements while the sixth period contains 32 elements? Relate this to electronic configuration. | For the first period, the principal quantum number is $$n=1$$. This shell contains only one subshell, the 1s subshell. The 1s subshell has one orbital, which can hold a maximum of 2 electrons. This corresponds to the two elements in the first period: Hydrogen ($$1s^1$$) and Helium ($$1s^2$$). For the sixth period, the ... | Chemistry 11th |
How can you predict the period and group of an element from its electronic configuration? Apply this to an element with configuration $[Ar]3d^{10}4s^2 4p^3$. | The given electronic configuration is $$[Ar]3d^{10}4s^2 4p^3$$. The period is the highest principal quantum number ($$n$$). The group number for a p-block element is $$10 + (\text{number of valence electrons})$$. The highest principal quantum number is 4. The number of valence electrons in the $$n=4$$ shell ($$4s^2$$ a... | Chemistry 11th |
What are $d$-block elements? Why are they also called transition elements? | The d-block elements are elements in the periodic table where the last electron, the differentiating electron, enters the d-orbital of the penultimate shell. Their general outer electron configuration is given by $$(n-1)d^{1-10} ns^{0-2}$$, where $$n$$ is the outermost shell. They are called transition elements because... | Chemistry 11th |
How does the ionic radius of a cation compare with the atomic radius of its parent atom? Explain with an example. | A cation is formed when a neutral atom loses one or more of its valence electrons. The loss of electrons often means the entire outermost electron shell is removed, causing a significant decrease in radius. After losing electrons, there are more protons than electrons. This increases the effective nuclear charge on the... | Chemistry 11th |
Why is the second ionization enthalpy of an element always higher than the first? Why is there a large jump in ionization enthalpy after removing valence electrons? | After one electron is removed, a positive ion (cation) is formed. The number of protons is now greater than the number of electrons. This imbalance increases the effective nuclear charge ($$Z_{eff}$$) experienced by the remaining electrons. The nucleus pulls the electron cloud in more tightly. Therefore, more energy is... | Chemistry 11th |
Why is the second electron gain enthalpy of oxygen positive while the first is negative? | The first electron gain enthalpy involves a neutral oxygen atom gaining an electron: $$\mathrm{O}(g) + e^- \rightarrow \mathrm{O}^{-}(g)$$. The attraction between the oxygen nucleus and the incoming electron causes a release of energy, so the value is negative (exothermic). The second electron gain enthalpy involves a ... | Chemistry 11th |
Why do elements of group 14 show a valence of 4 while elements of group 17 show valence of 1? | Valence is the combining capacity of an element, determined by the number of electrons an atom needs to lose, gain, or share to achieve a stable electron configuration (usually an octet). Elements in Group 14, like Carbon ($$C$$), have 4 valence electrons in their outermost shell. To achieve a stable octet of 8 electro... | Chemistry 11th |
What are amphoteric oxides? Give two examples and show their reactions with acids and bases. | Amphoteric oxides are metal oxides that react with both acids and bases to produce salt and water. Example 1: Zinc Oxide ($$ZnO$$). It reacts with acid (HCl) and base (NaOH). Reaction with acid: $$ZnO + 2HCl \rightarrow ZnCl_2 + H_2O$$. Reaction with base: $$ZnO + 2NaOH \rightarrow Na_2ZnO_2 + H_2O$$. Example 2: Alumin... | Chemistry 11th |
The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of $1,368 \mathrm{kHz}$ (kilo hertz). Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to? | The given frequency of the radio wave is $$\nu = 1,368 \text{ kHz}$$ or $$1.368 \times 10^6 \text{ Hz}$$. The speed of light is $$c = 3.0 \times 10^8 \text{ m/s}$$. The relationship between wavelength ($$\lambda$$), frequency ($$\nu$$), and the speed of light ($$c$$) is given by the formula: $$\lambda = \frac{c}{\nu}$$... | Chemistry 11th |
Arrange the following in increasing order of metallic character: Si, Be, Mg, Na, P. Justify your answer. | The given elements are Silicon (Si), Beryllium (Be), Magnesium (Mg), Sodium (Na), and Phosphorus (P). Metallic character decreases as we move from left to right across a period and increases as we move down a group in the periodic table. Na, Mg, Si, and P are in Period 3. Metallic character decreases across the period:... | Chemistry 11th |
Why do Li and Mg show similar properties despite being in different groups? List two similarities between them. | Lithium and Magnesium show a diagonal relationship in the periodic table. This relationship exists because they have a similar charge-to-radius ratio (charge density). Moving from Li across the period to Be, the ionic radius decreases. Moving down the group from Be to Mg, the ionic radius increases. These two effects p... | Chemistry 11th |
\
(i) Calculate the total number of electrons present in one mole of methane.\
(ii) Find\
(a) the total number and\
(b) the total mass of neutrons in 7 mg of 14C. (Assume that mass of a neutron $=1.675 \times 10-27 \mathrm{~kg}$ ).\
(iii) Find\
(a) the total number and\
(b) the total mass of protons in 34 mg of $\mathr... | **Part (i): Electrons in Methane ($$CH_4$$)**
A molecule of methane contains $$6 + 4(1) = 10$$ electrons. One mole of methane contains $$10 \times N_A = 10 \times 6.022 \times 10^{23} = 6.022 \times 10^{24}$$ electrons. **Part (ii): Neutrons in Carbon-14 ($$^{14}C$$)**
The number of moles in 7 mg of $$^{14}C$$ is $$\fr... | Chemistry 11th |
The work function for caesium atom is 1.9 eV . Calculate\
(a) the threshold wavelength and\
(b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm , calculate the kinetic energy and the velocity of the ejected photoelectron. | Given: Work function $$\phi = 1.9 \text{ eV}$$ and incident wavelength $$\lambda = 500 \text{ nm}$$. We convert the work function to Joules: $$\phi = 1.9 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 3.04 \times 10^{-19} \text{ J}$$. The threshold wavelength ($$\lambda_0$$) is found using the work function for... | Chemistry 11th |
If the photon of the wavelength 150 pm strikes an atom and one of tis inner bound electrons is ejected out with a velocity of $1.5 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}$, calculate the energy with which it is bound to the nucleus. | The given data are: Wavelength of the photon, $$\lambda = 150 \text{ pm} = 150 \times 10^{-12} \text{ m}$$. Velocity of the ejected electron, $$v = 1.5 \times 10^7 \text{ m/s}$$. We also use the constants: Planck's constant, $$h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s}$$; speed of light, $$c = 3.0 \times 10^8 \... | Chemistry 11th |
Emission transitions in the Paschen series end at orbit $\mathrm{n}=3$ and start from orbit n and can be represeted as $v=3.29 \times 10^{15}(\mathrm{~Hz})\left[1 / 3^{2}-1 / \mathrm{n}^{2}\right]$
Calculate the value of n if the transition is observed at 1285 nm . Find the region of the spectrum. | Given: Wavelength $$\lambda = 1285 \text{ nm} = 1285 \times 10^{-9} \text{ m}$$. The frequency is given by $$v = 3.29 \times 10^{15} (\text{Hz}) [1/3^2 - 1/n^2]$$. First, calculate the frequency using the relationship between the speed of light ($$c$$), frequency ($$v$$), and wavelength ($$\lambda$$): $$v = \frac{c}{\l... | Chemistry 11th |
Calculate the number of protons, neutrons and electrons in ${ }_{35}^{80} \mathrm{Br}$. | Given the notation $$_{35}^{80}\mathrm{Br}$$, the atomic number ($$Z$$) is 35 and the mass number ($$A$$) is 80. The number of protons is equal to the atomic number. The number of electrons is equal to the number of protons in a neutral atom. Number of protons = $$Z = 35$$. Number of electrons = 35. The number of neutr... | Chemistry 11th |
The wavelength range of the visible spectrum extends from violet ( 400 nm ) to red $(750 \mathrm{~nm})$. Express these wavelengths in frequencies (Hz). ( $1 \mathrm{~nm}=10^{-9} \mathrm{~m}$ ) | Given the wavelengths for violet light ($$\lambda_{violet}$$) as 400 nm and red light ($$\lambda_{red}$$) as 750 nm. The speed of light ($$c$$) is $$3.0 \times 10^8$$ m/s. We convert nanometers to meters: $$400 \times 10^{-9}$$ m and $$750 \times 10^{-9}$$ m. The relationship between speed of light ($$c$$), frequency (... | Chemistry 11th |
Calculate\
(a) wavenumber and\
(b) frequency of yellow radiation having wavelength 5800 Å. | The given wavelength is $$\lambda = 5800$$ Angstrom. This is converted to meters: $$\lambda = 5800 \times 10^{-10}$$ m. The speed of light is $$c = 3 \times 10^8$$ m/s. The formula for wavenumber is $$\bar{\nu} = \frac{1}{\lambda}$$. The formula for frequency is $$\nu = \frac{c}{\lambda}$$. For wavenumber: $$\bar{\nu... | Chemistry 11th |
Calculate energy of one mole of photons of radiation whose frequency is $5 \times 10^{14}$ Hz . | Given frequency ($$\nu$$) is $$5 \times 10^{14}$$ Hz. We need the energy for one mole, which is $$6.022 \times 10^{23}$$ photons. Planck's constant ($$h$$) is $$6.626 \times 10^{-34}$$ J$\cdot$s. The energy of a single photon is given by Planck's equation, $$E = h\nu$$. To find the energy for one mole of photons, we mu... | Chemistry 11th |
A 100 watt bulb emits monochromatic light of wavelength 400 nm . Calculate the number of photons emitted per second by the bulb. | Given: Power of the bulb $$P = 100$$ watts, and wavelength of light $$\lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m}$$. We also use Planck's constant $$h = 6.626 \times 10^{-34} \text{ J s}$$ and the speed of light $$c = 3 \times 10^8 \text{ m/s}$$. The total number of photons emitted per second ($$n$$) is the... | Chemistry 11th |
Calculate the energy associated with the first orbit of $\mathrm{He}^{+}$. What is the radius of this orbit? | The given atom is a Helium ion ($$\mathrm{He}^{+}$$), which is a hydrogen-like atom. The atomic number ($$Z$$) for Helium is 2. We are calculating for the first orbit, so the principal quantum number ($$n$$) is 1. The formula for the energy of an electron in the $$n^{th}$$ orbit of a hydrogen-like atom is $$E_n = -13.6... | Chemistry 11th |
What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of $10 \mathrm{~m} \mathrm{~s}^{-1}$ ? | Given: Mass ($$m$$) = 0.1 kg, Velocity ($$v$$) = 10 m/s. Planck's constant ($$h$$) = $$6.626 \times 10^{-34}$$ J s. The de Broglie wavelength formula is: $$\lambda = \frac{h}{mv}$$ Substitute the values into the formula: $$\lambda = \frac{6.626 \times 10^{-34} \text{ J s}}{(0.1 \text{ kg})(10 \text{ m/s})} = 6.626 \tim... | Chemistry 11th |
The mass of an electron is $9.1 \times 10^{-31} \mathrm{~kg}$. If its K.E. is $3.0 \times 10^{-25} \mathrm{~J}$, calculate its wavelength. | Given data: Mass $$m = 9.1 \times 10^{-31} \text{ kg}$$, Kinetic Energy $$K.E. = 3.0 \times 10^{-25} \text{ J}$$. Planck's constant $$h = 6.626 \times 10^{-34} \text{ J s}$$. The de Broglie wavelength $$\lambda$$ is related to kinetic energy by the formula: $$\lambda = \frac{h}{\sqrt{2m(K.E.)}}$$. Substitute the values... | Chemistry 11th |
What is the total number of orbitals associated with the principal quantum number $n=3$ ? | The given principal quantum number is $$n=3$$. The total number of orbitals for a given principal quantum number $$n$$ is calculated by the formula $$n^2$$. Substituting the given value, we get $$3^2 = 9$$. Therefore, there are 9 orbitals associated with the principal quantum number $$n=3$$. **Final Answer:** The total... | Chemistry 11th |
Using $s, p, d, f$ notations, describe the orbital with the following quantum numbers\
(a) $n=2, l=1$,\
(b) $n=4, l=0$,\
(c) $n=5$, $l=3$,\
(d) $n=3, l=2$ | The principal quantum number, $$n$$, gives the numerical designation for the orbital. The azimuthal quantum number, $$l$$, determines the letter designation for the orbital: $$l=0$$ is s, $$l=1$$ is p, $$l=2$$ is d, and $$l=3$$ is f. Combine the number from $$n$$ and the letter from $$l$$ to write the orbital notation.... | Chemistry 11th |
\
(i) Calculate the number of electrons which will together weigh one gram.\
(ii) Calculate the mass and charge of one mole of electrons. | Given: Mass of an electron, $$m_e = 9.11 \times 10^{-31}$$ kg. Target mass = 1 g or $$10^{-3}$$ kg. The number of electrons $$n$$ is the total mass divided by the mass of a single electron: $$n = \frac{\text{Total Mass}}{m_e}$$. $$n = \frac{10^{-3} \text{ kg}}{9.11 \times 10^{-31} \text{ kg}} = 1.098 \times 10^{27}$$ e... | Chemistry 11th |
Write the complete symbol for the atom with the given atomic number ( $Z$ ) and atomic mass (A)\
(i) $\mathrm{Z}=17, \mathrm{~A}=35$.\
(ii) $\mathrm{Z}=92, \mathrm{~A}=233$.\
(iii) $\mathrm{Z}=4, \mathrm{~A}=9$. | The symbol for an atom is written as $$^{A}_{Z}X$$, where $$A$$ is the mass number, $$Z$$ is the atomic number, and $$X$$ is the element's symbol. For (i), $$Z=17$$ corresponds to the element Chlorine (Cl). The symbol is $$^{35}_{17}Cl$$. For (ii), $$Z=92$$ corresponds to the element Uranium (U). The symbol is $$^{233}... | Chemistry 11th |
Yellow light emitted from a sodium lamp has a wavelength ( $\lambda$ ) of 580 nm . Calculate the frequency ( $v$ ) and wavenumber ( $\bar{v}$ ) of the yellow light. | The given wavelength is $$\lambda = 580$$ nm. We convert this to meters: $$580 \times 10^{-9}$$ m. The speed of light is $$c = 3.0 \times 10^8$$ m/s. The formula for frequency is $$v = \frac{c}{\lambda}$$. The formula for wavenumber is $$\bar{v} = \frac{1}{\lambda}$$. Frequency calculation: $$v = \frac{3.0 \times 10^8 ... | Chemistry 11th |
Find energy of each of the photons which\
(i) correspond to light of frequency $3 \times 1015 \mathrm{~Hz}$.\
(ii) have wavelength of $0.50 \AA$. | Given data: For (i), frequency $$v = 3 \times 10^{15}$$ Hz. For (ii), wavelength $$\lambda = 0.50$$ Angstrom, which is $$0.50 \times 10^{-10}$$ m. We use Planck's constant $$h = 6.626 \times 10^{-34}$$ J s and the speed of light $$c = 3 \times 10^8$$ m/s. Formulas used: For (i), the energy $$E$$ is calculated using $$... | Chemistry 11th |
A photon of wavelength $4 \times 10^{-7} \mathrm{~m}$ strikes on metal surface, the work function of the metal being 2.13 eV . Calculate\
(i) the energy of the photon (eV),\
(ii) the kinetic energy of the emission, and\
(iii) the velocity of the photoelectron ( $1 \mathrm{eV}=1.6020 \times 10^{-19} \mathrm{~J}$ ). | Given: Wavelength ($$\lambda$$) = $$4 \times 10^{-7}$$ m, Work function ($$\phi$$) = 2.13 eV. Constants: Planck's constant ($$h$$) = $$6.626 \times 10^{-34}$$ J$\cdot$s, Speed of light ($$c$$) = $$3 \times 10^8$$ m/s, Mass of electron ($$m_e$$) = $$9.1 \times 10^{-31}$$ kg. The energy of a photon is given by the formul... | Chemistry 11th |
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm . Name the series to which this transition belongs and the region of the spectrum. | The given data for the hydrogen atom is: initial radius $$r_i = 1.3225$$ nm and final radius $$r_f = 211.6$$ pm. First, convert radii to the same unit (pm): $$r_i = 1322.5$$ pm. The principal quantum numbers ($$n$$) for the initial and final orbits are found using the Bohr radius formula, $$r_n = n^2 \times 52.9$$ pm. ... | Chemistry 11th |
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in $\mathrm{kJ} \mathrm{mol}^{-1}$. | Given wavelength $$\lambda = 242 \ \text{nm}$$ which is $$242 \times 10^{-9} \ \text{m}$$. We also need Planck's constant $$h = 6.626 \times 10^{-34} \ \text{J s}$$, the speed of light $$c = 3.00 \times 10^8 \ \text{m s}^{-1}$$, and Avogadro's number $$N_A = 6.022 \times 10^{23} \ \text{mol}^{-1}$$. The energy ($$E$$) ... | Chemistry 11th |
A 25 watt bulb emits monochromatic yellow light of wavelength of $0.57 \mu \mathrm{~m}$. Calculate the rate of emission of quanta per second. | Given: Power $$P = 25$$ W (or 25 J/s), and Wavelength $$\lambda = 0.57 \mu\text{m} = 0.57 \times 10^{-6}$$ m. We use Planck's constant $$h = 6.626 \times 10^{-34}$$ J$\cdot$s and the speed of light $$c = 3 \times 10^8$$ m/s. The energy ($$E$$) of a single quantum is found using the formula $$E = \frac{hc}{\lambda}$$. T... | Chemistry 11th |
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength $6800 \AA$. Calculate threshold frequency ( $v_{0}$ ) and work function $\left(\mathrm{W}_{0}\right)$ of the metal. | The radiation wavelength $$\lambda$$ is 6800 Angstrom. Since the emitted electrons have zero velocity, this is the threshold wavelength, so $$\lambda_0 = 6800 \text{ Angstrom} = 6800 \times 10^{-10}$$ m. The speed of light is $$c = 3 \times 10^8$$ m/s. Planck's constant is $$h = 6.626 \times 10^{-34}$$ J$\cdot$s. The... | Chemistry 11th |
How much energy is required to ionise a H atom if the electron occupies $n=5$ orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from $n=1$ orbit). | The initial orbit is $$n_i = 5$$ and the final state is $$n_f = \infty$$ for ionization. The energy of an electron in a hydrogen atom is given by the formula: $$E_n = -2.18 \times 10^{-18} \left( \frac{1}{n^2} \right) \text{ J}$$. The energy for ionization from $$n=5$$ is $$\Delta E = E_\infty - E_5 = 0 - \left( -2.18 ... | Chemistry 11th |
Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen. | For the Balmer series, the final principal quantum number is $$n_1 = 2$$. The longest wavelength (lowest energy) transition occurs from the next level up, so the initial principal quantum number is $$n_2 = 3$$. The Rydberg constant $$R_H$$ is 109677 cm-»^1. The Rydberg formula is used to calculate the wavenumber ($$\t... | Chemistry 11th |
What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is $-2.18 \times 10^{-11} \mathrm{ergs}$. | The ground state energy is given as $$E_1 = -2.18 \times 10^{-11} \mathrm{ergs}$$. First, convert this energy to Joules: $$E_1 = -2.18 \times 10^{-11} \mathrm{ergs} \times (10^{-7} \mathrm{J/erg}) = -2.18 \times 10^{-18} \mathrm{J}$$. The energy of an electron in a specific orbit $$n$$ is given by the formula $$E_n = E... | Chemistry 11th |
Calculate the wavelength of an electron moving with a velocity of $2.05 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}$. | The given information is the velocity of the electron, $$v = 2.05 \times 10^{7} \text{ m s}^{-1}$$. We also use the known mass of an electron, $$m = 9.11 \times 10^{-31} \text{ kg}$$, and Planck's constant, $$h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s}$$. The de Broglie equation is used to find the wavelength of... | Chemistry 11th |
Which of the following are isoelectronic species i.e., those having the same number of electrons?
$$
\mathrm{Na}^{+}, \mathrm{K}^{+}, \mathrm{Mg}^{2+}, \mathrm{Ca}^{2+}, \mathrm{S}^{2-}, \mathrm{Ar} .
$$ | To find the number of electrons in an ion, we use its atomic number (Z). For a neutral atom, the number of electrons equals Z. For a cation, the number of electrons is Z - (charge). For an anion, the number of electrons is Z + (charge). Number of electrons in $$\mathrm{Na}^{+}$$ (Z=11) is $$11 - 1 = 10$$. Number of ele... | Chemistry 11th |
What is the lowest value of $n$ that allows $g$ orbitals to exist? | We need to find the lowest principal quantum number, $$n$$, for a $$g$$ orbital. The angular momentum quantum number, $$l$$, ranges from $$0$$ to $$n-1$$. The value of $$l$$ corresponds to orbital types: $$s=0$$, $$p=1$$, $$d=2$$, $$f=3$$, $$g=4$$. For a $$g$$ orbital to exist, the value of $$l$$ must be 4. According t... | Chemistry 11th |
An electron is in one of the $3 d$ orbitals. Give the possible values of $n, l$ and $m_{l}$ for this electron. | The electron is in a $$3d$$ orbital. The principal quantum number $$n$$ is the number in the orbital's name. The azimuthal quantum number $$l$$ is determined by the letter (s=0, p=1, d=2). The magnetic quantum number $$m_l$$ ranges from $$-l$$ to $$+l$$. For a $$3d$$ orbital, $$n=3$$. The letter 'd' corresponds to $$l=... | Chemistry 11th |
Give the number of electrons in the species $\mathrm{H}_{2}^{+}, \mathrm{H}_{2}$ and $\mathrm{O}_{2}^{+}$ | The species given are $$\mathrm{H}_{2}^{+}$$, $$\mathrm{H}_{2}$$, and $$\mathrm{O}_{2}^{+}$$. A neutral hydrogen atom has 1 electron, and a neutral oxygen atom has 8 electrons. The total number of electrons is the sum of the electrons from each constituent atom, adjusted for the overall charge of the species. $$ \text{... | Chemistry 11th |
\
(i) An atomic orbital has $\mathrm{n}=3$. What are the possible values of 1 and ml ?\
(ii) List the quantum numbers ( $m l$ and $l$ ) of electrons for $3 d$ orbital.\
(iii) Which of the following orbitals are possible?
$1 p, 2 s, 2 p$ and $3 f$ | For a principal quantum number $$n=3$$, the azimuthal quantum number $$l$$ can have integer values from $$0$$ to $$n-1$$. The magnetic quantum number $$m_l$$ can have integer values from $$-l$$ to $$+l$$. For $$n=3$$, the possible values of $$l$$ are $$0, 1, 2$$. When $$l=0$$, $$m_l=0$$. When $$l=1$$, $$m_l = -1, 0, +1... | Chemistry 11th |
Using $s, p, d$ notations, describe the orbital with the following quantum numbers.\
(a) $n=1, l=0$;\
(b) $n=3 ; l=1$\
(c) $n=4 ; l=2$;\
(d) $n=4 ; l=3$. | The principal quantum number $$n$$ specifies the shell. The angular momentum quantum number $$l$$ specifies the subshell. The value of $$l$$ corresponds to a specific letter designation for the subshell: $$l=0$$ is s, $$l=1$$ is p, $$l=2$$ is d, and $$l=3$$ is f. Combine the principal quantum number $$n$$ with the lett... | Chemistry 11th |
Explain, giving reasons, which of the following sets of quantum numbers are not possible.\
(a) $n=0$,
$l=0$,
$m_{l}=0$,
$m_{s}=+\frac{1}{2}$\
(b) $\mathrm{n}=1$,
$\mathrm{ml}=0$,
$\mathrm{ms}=-\frac{1}{2}$\
(c) $\mathrm{n}=1$,
$1=0$,
$m_{l}=0$,
$m_{s}=+\frac{1}{2}$\
(d) $n=2$,
$l=1$,
$m_{l}=0$,
$m_{s}=-\frac{1}{2}$\
(e... | The principal quantum number, $$n$$, must be a positive integer ($$n=1, 2, 3, ...$$). In set (a), $$n=0$$, which is not allowed. The angular momentum quantum number, $$l$$, is not specified in set (b). A complete set of quantum numbers requires a value for $$l$$. The angular momentum quantum number, $$l$$, must be in t... | Chemistry 11th |
Calculate the energy required for the process
$\mathrm{He}^{+}(\mathrm{g}) \rightarrow \mathrm{He}^{2+}(\mathrm{g})+\mathrm{e}^{-}$
The ionization energy for the H atom in the ground state is $2.18 \times 10^{-18} \mathrm{~J}$ atom $^{-1}$ | The species $$He^{+}$$ is a hydrogen-like ion because it has only one electron. Its atomic number, $$Z$$, is 2. The given ionization energy for the H atom is $$2.18 \times 10^{-18}$$ J/atom, where $$Z=1$$. The formula for the ionization energy ($$IE$$) of a hydrogen-like species is $$IE_{\text{species}} = IE_H \times Z... | Chemistry 11th |
If the diameter of a carbon atom is 0.15 nm , calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.
$2.362 \times 10^{8}$ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.... | Given: Diameter of carbon atom $$d = 0.15 ext{ nm} = 0.15 imes 10^{-7} ext{ cm}$$. Length of scale $$L = 20 ext{ cm}$$. For the second arrangement: Number of atoms $$N = 2.362 imes 10^8$$ and total length $$L_2 = 2.4 ext{ cm}$$.
**Part 1: Number of atoms in 20 cm**
The number of atoms $$N = rac{ ext{Total Leng... | Chemistry 11th |
$2 \times 10^{8}$ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm . | The number of carbon atoms is $$n = 2 \times 10^{8}$$. The total length of the arrangement is $$L = 2.4$$ cm. The total length is the number of atoms multiplied by the diameter ($$2r$$) of each atom. The formula is $$L = n \times 2r$$. To find the radius, we rearrange the formula to $$r = \frac{L}{2n}$$. Substitute the... | Chemistry 11th |
A certain particle carries $2.5 \times 10^{-16} \mathrm{C}$ of static electric charge. Calculate the number of electrons present in it. | The total charge is $$q = 2.5 \times 10^{-16} \mathrm{C}$$. The charge of a single electron is $$e = 1.602 \times 10^{-19} \mathrm{C}$$. The formula for the quantization of charge is $$q = ne$$, where $$n$$ is the number of electrons. This is rearranged to find $$n$$: $$n = \frac{q}{e}$$. Substituting the given values:... | Chemistry 11th |
In Milikan's experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is $-1.282 \times 10^{-18} \mathrm{C}$, calculate the number of electrons present on it. | The total charge on the oil drop is $$q = -1.282 \times 10^{-18}$$ C. The charge of a single electron is $$e = -1.602 \times 10^{-19}$$ C. The formula for the quantization of charge is $$q = ne$$. To find the number of electrons, $$n$$, we rearrange the formula to $$n = \frac{q}{e}$$. Substituting the given values: $$n... | Chemistry 11th |
In Rutherford's experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the $\alpha$-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results? | The scattering of $$\alpha$$-particles is due to the electrostatic force of repulsion from the positively charged nucleus. The magnitude of this force depends on the charge of the nucleus. Heavy atoms like gold ($$Z=79$$) have a large, highly concentrated positive charge in their nucleus. Light atoms like aluminum ($$Z... | Chemistry 11th |
An ion with mass number 37 possesses one unit of negative charge. If the ion conatins $11.1 \%$ more neutrons than the electrons, find the symbol of the ion. | The mass number ($$A$$) is 37. The charge is -1. Let the number of electrons be $$e$$ and neutrons be $$n$$. The number of neutrons is 11.1% more than the electrons, so $$n = e + 0.111e = 1.111e$$. The fundamental equations are: Mass Number $$A = \text{protons} + \text{neutrons}$$, and Electrons $$e = \text{protons} - ... | Chemistry 11th |
Arrange the following type of radiations in increasing order of frequency:\
(a) radiation from microwave oven\
(b) amber light from traffic signal\
(c) radiation from FM radio\
(d) cosmic rays from outer space and\
(e) X-rays. | The different types of radiation given are: (a) Microwaves, (b) Amber light (Visible), (c) FM radio waves, (d) Cosmic rays (Gamma rays), and (e) X-rays. The electromagnetic spectrum is arranged in order of increasing frequency as follows: Radio waves < Microwaves < Infrared < Visible light < Ultraviolet < X-rays < Gamm... | Chemistry 11th |
Nitrogen laser produces a radiation at a wavelength of 337.1 nm . If the number of photons emitted is $5.6 \times 10^{24}$, calculate the power of this laser. | The given wavelength is $$ \lambda = 337.1 \ \text{nm} = 337.1 \times 10^{-9} \ \text{m} $$. The number of photons emitted is $$ n = 5.6 \times 10^{24} $$. We use Planck's constant $$ h = 6.626 \times 10^{-34} \ \text{J s} $$ and the speed of light $$ c = 3 \times 10^{8} \ \text{m/s} $$. The total energy is the number ... | Chemistry 11th |
In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of $3.15 \times 10^{-18} \mathrm{~J}$ from the radiations of 600 nm , calculate the number of photons received by the detector. | The given total energy is $$E_{total} = 3.15 \times 10^{-18} \mathrm{~J}$$ and the wavelength is $$\lambda = 600 \mathrm{~nm}$$ or $$600 \times 10^{-9} \mathrm{~m}$$. Planck's constant is $$h = 6.626 \times 10^{-34} \mathrm{~J \cdot s}$$ and the speed of light is $$c = 3 \times 10^8 \mathrm{~m/s}$$. The energy of a sin... | Chemistry 11th |
Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is $2.5 \times 10^{15}$, calculate the energy of the source. | Given data: Number of photons $$N = 2.5 \times 10^{15}$$. Time duration $$t = 2 \text{ ns} = 2 \times 10^{-9} \text{ s}$$. Planck's constant $$h = 6.626 \times 10^{-34} \text{ J s}$$. The energy of the source is the total energy of all emitted photons, given by the formula $$E = N h \nu$$. The frequency ($$\nu$$) of th... | Chemistry 11th |
Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is $1.6 \times 10^{6} \mathrm{ms}^{-1}$, calculate de Broglie wavelength associate... | Given the velocity of the electron, $$v = 1.6 \times 10^{6} \mathrm{ms}^{-1}$$. The mass of an electron is $$m = 9.11 \times 10^{-31}$$ kg, and Planck's constant is $$h = 6.626 \times 10^{-34}$$ J s. The de Broglie wavelength is calculated using the formula: $$\lambda = \frac{h}{mv}$$. Substituting the values into the ... | Chemistry 11th |
Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm , calculate the characteristic velocity associated with the neutron. | Given: Wavelength $$\lambda = 800 \text{ pm} = 800 \times 10^{-12} \text{ m}$$. The mass of a neutron $$m = 1.67493 \times 10^{-27} \text{ kg}$$. Planck's constant $$h = 6.626 \times 10^{-34} \text{ J s}$$. The de Broglie equation relates wavelength and momentum: $$\lambda = \frac{h}{mv}$$. We rearrange for velocity: $... | Chemistry 11th |
If the velocity of the electron in Bohr's first orbit is $2.19 \times 10^{6} \mathrm{~ms}^{-1}$, calculate the de Broglie wavelength associated with it. | The given velocity of the electron is $$v = 2.19 \times 10^{6} \mathrm{~ms}^{-1}$$. We also use the mass of an electron, $$m = 9.11 \times 10^{-31} \mathrm{~kg}$$, and Planck's constant, $$h = 6.626 \times 10^{-34} \mathrm{~J \cdot s}$$. The de Broglie wavelength ($$\lambda$$) is calculated using the formula: $$\lambda... | Chemistry 11th |
The velocity associated with a proton moving in a potential difference of 1000 V is $4.37 \times 10^{5} \mathrm{~ms}^{-1}$. If the hockey ball of mass 0.1 kg is moving with this velocity, calcualte the wavelength associated with this velocity. | Mass of the hockey ball ($$m$$) = 0.1 kg. Velocity ($$v$$) = $$4.37 \times 10^{5} \mathrm{~ms}^{-1}$$. Planck's constant ($$h$$) = $$6.626 \times 10^{-34} \mathrm{~J \cdot s}$$. The de Broglie wavelength equation is $$\lambda = \frac{h}{mv}$$. Substituting the values: $$\lambda = \frac{6.626 \times 10^{-34} \mathrm{~J ... | Chemistry 11th |
Indicate the number of unpaired electrons in :\
(a) P,\
(b) Si,\
(c) Cr,\
(d) Fe and\
(e) Kr. | First, write the ground-state electron configuration for each element based on its atomic number. Note any exceptions to the Aufbau principle, such as for Chromium. Next, apply Hund's rule to the outermost, partially filled subshells. This rule states that electrons will fill empty orbitals within a subshell before pai... | Chemistry 11th |
\
(a) How many subshells are associated with $n=4$ ?\
(b) How many electrons will be present in the subshells having $\mathrm{m}_{\mathrm{s}}$ value of $-1 / 2$ for $n=4$ ?
\end{enumerate} | The principal quantum number is given as $$n=4$$. For a given $$n$$, the number of subshells is equal to $$n$$. The values of the azimuthal quantum number $$l$$ range from $$0$$ to $$n-1$$. For $$n=4$$, the possible values of $$l$$ are $$0, 1, 2, 3$$. These correspond to the s, p, d, and f subshells, respectively. This... | Chemistry 11th |
How does the classification of elements help in studying their properties and predicting the behavior of new elements? | The classification of elements is based on the Periodic Law, which states that the properties of elements are periodic functions of their atomic number. Elements are arranged in the periodic table in groups (vertical columns) and periods (horizontal rows). Elements within the same group share similar valence electron c... | Chemistry 11th |
What was the total number of elements known in 1800, 1865, and at present? How did the increasing number of elements necessitate their classification? | In 1800, about 31 elements were known. By 1865, this number had increased to 63 elements. At present, there are 118 known elements. The growing number of elements and their compounds made it difficult for scientists to study and remember their individual chemical and physical properties. Scientists began to search for ... | Chemistry 11th |
In Dobereiner's Triads, how is the atomic weight of the middle element related to the atomic weights of the other two elements? Verify this relationship using the Li-Na-K triad. | The given triad is Lithium (Li), Sodium (Na), and Potassium (K). The atomic weight of Li is 6.9 u, Na is 23.0 u, and K is 39.1 u. The atomic weight of the middle element is the arithmetic mean of the other two elements. $$W_{Na} = \frac{W_{Li} + W_{K}}{2}$$. Substitute the atomic weights into the formula: $$W_{Na} = \f... | Chemistry 11th |
State Newlands' Law of Octaves. Why was it named after the musical scale? | John Newlands arranged the known elements in order of increasing atomic weight. He observed that the properties of every eighth element were similar to the first one. This periodic repetition was named the 'Law of Octaves' due to its similarity to the seven notes of the musical scale, where the eighth note is a repetit... | Chemistry 11th |
What were the limitations of Newlands' Law of Octaves? Up to which element did this law seem to hold true? | The law was only applicable up to Calcium. It was not valid for elements of higher atomic masses. The discovery of noble gases disrupted the pattern, as they could not be accommodated in his table. Newlands placed some unlike elements in the same slot to fit his pattern. For example, he placed cobalt and nickel in the ... | Chemistry 11th |
What were Eka-aluminium and Eka-silicon? How did Mendeleev's predictions for these elements compare with the actual properties of gallium and germanium? | Dmitri Mendeleev left gaps in his periodic table for elements that he hypothesized existed but had not yet been discovered. He used the prefix "Eka" (from Sanskrit for "one") to name these predicted elements based on their position below a known element in the same group. Eka-aluminium was the name given to the predict... | Chemistry 11th |
Why did Mendeleev leave gaps in his periodic table? How did this demonstrate the strength of his classification system? | Mendeleev arranged the known elements in order of increasing atomic mass, grouping them by similar chemical properties. When the properties of the next known element did not fit the pattern of a group, he left a gap. He hypothesized these gaps represented elements that had not yet been discovered. He used the propertie... | Chemistry 11th |
State the Modern Periodic Law. How does it differ from Mendeleev's Periodic Law? | Mendeleev's Periodic Law states that the physical and chemical properties of the elements are a periodic function of their atomic masses. The Modern Periodic Law states that the physical and chemical properties of the elements are a periodic function of their atomic number. The fundamental difference is the basis for o... | Chemistry 11th |
Who discovered that atomic number is a more fundamental property than atomic mass? What experimental evidence led to this conclusion? | In 1913, Henry Moseley studied the characteristic X-rays emitted by different chemical elements when they were bombarded with high-energy electrons. He discovered a precise mathematical relationship between the frequency of the emitted X-rays ($$\nu$$) and the atomic number ($$Z$$) of the element. The relationship, kno... | Chemistry 11th |
Why is the Modern Periodic Law based on atomic number rather than atomic mass considered to be a more accurate basis for classification? | Mendeleev's periodic law stated that the properties of elements are a periodic function of their atomic masses. This model faced issues with 'anomalous pairs' like Argon ($$39.9 \text{ u}$$) and Potassium ($$39.1 \text{ u}$$), where the element with the higher mass appeared first. The existence of isotopes also created... | Chemistry 11th |
What are periods and groups in the modern periodic table? How many periods and groups are present in the long form of the periodic table? | The horizontal rows of elements in the periodic table are known as periods. Elements in the same period have the same number of electron shells. The vertical columns of elements in the periodic table are known as groups. Elements in the same group share the same number of valence electrons and thus have similar chemica... | Chemistry 11th |
How does the period number relate to the principal quantum number of the valence shell electrons? Explain with examples. | The periods of the periodic table are the horizontal rows. The principal quantum number, $$n$$, indicates the main energy level or shell of an atom's electrons. The period number is identical to the principal quantum number of the outermost electron shell (the valence shell). For example, Sodium (Na) is in Period 3. It... | Chemistry 11th |
What is the IUPAC system for naming elements with atomic number greater than 100? Give the roots used for digits 0-9. | The IUPAC system provides a temporary name for elements with atomic numbers greater than 100 based directly on their atomic number. Each digit in the atomic number corresponds to a specific numerical root. The roots are: 0=nil, 1=un, 2=bi, 3=tri, 4=quad, 5=pent, 6=hex, 7=sept, 8=oct, 9=enn. The roots are concatenated i... | Chemistry 11th |
What would be the IUPAC name and symbol for the element with atomic number 119? | The atomic number is 119. The IUPAC systematic naming convention uses roots for each digit: 1 (un), 1 (un), and 9 (enn). The name is formed by combining these roots and adding the suffix '-ium'. The symbol is the first letter of each root. Name: un + un + enn + ium = Ununennium. Symbol: U + u + e = Uue. The element wit... | Chemistry 11th |
Why was a systematic nomenclature for elements with $Z > 100$ necessary? What controversies led to this decision? | During the Cold War, research groups in the United States (at Berkeley) and the Soviet Union (at Dubna) began synthesizing new elements with $$Z > 100$$. Both groups often claimed discovery of the same element at nearly the same time, leading to disputes over who had the right to name it. This period is often called th... | Chemistry 11th |
Why do lanthanoids show very similar chemical properties? What is lanthanoid contraction? | The general electron configuration for lanthanoids is $$[\text{Xe}] 4f^{1-14} 5d^{0-1} 6s^2$$. The last electron enters the inner $$4f$$ orbital, while the outermost $$6s$$ orbital, which determines chemical properties, remains unchanged with two electrons. The $$4f$$ electrons have a very poor shielding effect. As the... | Chemistry 11th |
How does the electronic configuration of an element determine its position in the periodic table? Explain with the example of chlorine ($Z = 17$). | The atomic number ($$Z$$) of chlorine is 17, so it has 17 electrons. The electronic configuration is written by filling orbitals in order of increasing energy: $$1s^2 2s^2 2p^6 3s^2 3p^5$$. The period number is determined by the highest principal quantum number ($$n$$). For chlorine, the outermost electrons are in the ... | Chemistry 11th |
What are $s$-block elements? Write their general outer electronic configuration and list the groups included in this block. | The s-block elements of the periodic table are those elements in which the last electron enters the outermost s-orbital. The general outer electronic configuration for s-block elements is $$ns^{1-2}$$, where $$n$$ is the outermost principal energy level. For Group 1 (alkali metals), the configuration is $$ns^1$$. For G... | Chemistry 11th |
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