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Why are alkali metals highly reactive and electropositive? Relate this to their electronic configuration. | Alkali metals (Group 1) have a characteristic outer electronic configuration of $$ns^1$$, with a single electron in their valence shell. This lone valence electron is weakly bound and easily removed because of the shielding effect from the inner-shell electrons and its distance from the nucleus. This results in a very ... | Chemistry 11th |
What are the characteristic properties of $s$-block elements that distinguish them from other blocks? | The valence electrons of s-block elements occupy the s-orbital. Group 1 elements have an $$ns^1$$ configuration, and Group 2 elements have an $$ns^2$$ configuration. They possess low ionization enthalpies, meaning little energy is required to remove their outermost electron(s). This value decreases down each group. The... | Chemistry 11th |
What are $p$-block elements? Write their general outer electronic configuration and mention the range of groups in this block. | p-block elements are those in which the last electron enters any of the three p-orbitals of their respective outermost shells. The general outer electronic configuration for p-block elements is $$ns^2np^{1-6}$$, where $$n$$ is the principal quantum number. The p-block elements are located in Groups 13 to 18 of the peri... | Chemistry 11th |
Why is helium placed in the $p$-block along with noble gases despite having the electronic configuration $1s^2$? | Helium's electronic configuration is $$1s^2$$. Based on this, its last electron is in the s-orbital, suggesting it belongs to the s-block. The noble gases (Group 18) are defined by having a completely filled outermost electron shell. This configuration makes them exceptionally stable and chemically inert. Helium's oute... | Chemistry 11th |
Write the general outer electronic configuration of $d$-block elements. Why do these elements show variable oxidation states? | The general outer electronic configuration of d-block elements is $$(n-1)d^{1-10} ns^{1-2}$$. The d-block elements show variable oxidation states because the energy difference between the electrons in the $$(n-1)d$$ and $$ns$$ orbitals is very small. Due to this small energy gap, electrons from both the $$(n-1)d$$ and ... | Chemistry 11th |
What are $f$-block elements? Distinguish between lanthanoids and actinoids with respect to their electronic configuration. | The f-block elements are chemical elements in which the last electron enters the f-orbital of the ante-penultimate energy level, which is the (n-2) shell. They are also known as inner transition elements. Lanthanoids are the series of 14 elements following Lanthanum (from Cerium to Lutetium). In these elements, the dif... | Chemistry 11th |
What is atomic radius? How is it measured for (a) metallic elements and (b) non-metallic elements? | Atomic radius is the total distance from an atom-s nucleus to the outermost electron orbital. Since the edge of an orbital is not a well-defined boundary, the radius is determined by the distance between the nuclei of two bonded atoms. For metallic elements, the atomic radius is the metallic radius. It is defined as ... | Chemistry 11th |
How does atomic radius vary (a) across a period from left to right, and (b) down a group? Explain the reasons for these trends. | Atomic radius is typically defined as half the distance between the nuclei of two identical atoms bonded together. (a) Across a period (left to right), the atomic radius decreases. This is because the number of protons in the nucleus increases, but the number of electron shells remains the same. The stronger positive c... | Chemistry 11th |
Why is the ionic radius of an anion larger than the atomic radius of its parent atom? | An anion is formed when a neutral atom gains one or more electrons. The number of protons in the nucleus remains the same, so the attractive nuclear charge is constant. The added electrons increase the electron-electron repulsion in the outermost electron shell, pushing the electrons further apart. This increased repul... | Chemistry 11th |
Arrange the following species in increasing order of ionic radii: $\mathrm{Na}^+$, $\mathrm{Mg}^{2+}$, $\mathrm{Al}^{3+}$, $\mathrm{O}^{2-}$, $\mathrm{F}^-$. Justify your answer. | The given ions are $\mathrm{Al}^{3+}$, $\mathrm{Mg}^{2+}$, $\mathrm{Na}^+$, $\mathrm{F}^{-}$, and $\mathrm{O}^{2-}$. All of these ions have 10 electrons, making them an isoelectronic series. The number of protons are: $\mathrm{Al}^{3+}$ (13), $\mathrm{Mg}^{2+}$ (12), $\mathrm{Na}^+$ (11), $\mathrm{F}^{-}$ (9), $\mathrm... | Chemistry 11th |
What are isoelectronic species? Give examples of isoelectronic species with 10 electrons. | Isoelectronic species are atoms and ions that have the same number of electrons. The noble gas Neon ($$\mathrm{Ne}$$) has an atomic number of 10, so a neutral atom has 10 protons and 10 electrons. Anions are formed when atoms gain electrons. For example, $$\mathrm{O}^{2-}$$ (8 protons, 10 electrons) and $$\mathrm{F}^{-... | Chemistry 11th |
Among isoelectronic species, how does the ionic radius vary with nuclear charge? Explain with the example of $\mathrm{O}^{2-}$, $\mathrm{F}^-$, $\mathrm{Na}^+$, and $\mathrm{Mg}^{2+}$. | The given species are O^2-», F-», Na-º, and Mg^2-º. They are isoelectronic, each containing 10 electrons. For isoelectronic species, the ionic radius is inversely proportional to the nuclear charge ($$Z$$). A higher nuclear charge results in a stronger attraction between the nucleus and the electrons. The nuclear c... | Chemistry 11th |
Define ionization enthalpy. Why is it always a positive quantity? | Ionization enthalpy is the minimum energy required to remove the most loosely bound electron from an isolated neutral gaseous atom in its ground state. To remove an electron, energy must be supplied to overcome the electrostatic force of attraction between the positively charged nucleus and the negatively charged elect... | Chemistry 11th |
How does ionization enthalpy vary (a) across a period and (b) down a group? Explain the factors responsible for these trends. | Ionization enthalpy is the minimum energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. Across a period (left to right), ionization enthalpy generally increases. This is because the effective nuclear charge increases as electrons are added to the same principal en... | Chemistry 11th |
Explain why the first ionization enthalpy of boron is less than that of beryllium despite boron having a higher nuclear charge. | The electron configuration of Beryllium (Be) is $$1s^2 2s^2$$. The electron to be removed is from the $$2s$$ orbital. The electron configuration of Boron (B) is $$1s^2 2s^2 2p^1$$. The electron to be removed is from the $$2p$$ orbital. A $$2p$$ orbital is at a higher energy level than a $$2s$$ orbital. Therefore, the e... | Chemistry 11th |
The first ionization enthalpies of group 13 elements do not decrease regularly from B to Tl. Explain the anomaly observed for Ga and Tl. | The general trend for first ionization enthalpy is to decrease down a group due to increasing atomic size and shielding effect. Gallium (Ga) has a higher ionization enthalpy than Aluminum (Al) because Ga has ten 3d electrons in its inner shell. These d-electrons provide poor shielding for the outer 4p electron. The poo... | Chemistry 11th |
Define electron gain enthalpy. Why can it be either negative or positive? | Electron gain enthalpy is the enthalpy change when an isolated gaseous atom accepts an electron to form a monovalent anion. If the electron is added to an atom and energy is released, the process is exothermic. By convention, the electron gain enthalpy ($$\Delta_{eg}H$$) is given a negative sign. This is common for ele... | Chemistry 11th |
How does electron gain enthalpy vary across a period and down a group? Explain with examples. | Electron gain enthalpy ($$\Delta_{eg}H$$) is the energy change when an electron is added to an isolated gaseous atom. A more negative value indicates a greater affinity for an electron. Across a period (left to right), the effective nuclear charge increases and atomic size decreases. This results in a stronger attracti... | Chemistry 11th |
Why is the electron gain enthalpy of fluorine less negative than that of chlorine, even though F has a higher electronegativity? | Electron gain enthalpy is the energy released when an electron is added to a neutral atom in the gaseous state. A more negative value indicates a higher affinity for the electron. Fluorine is smaller and more electronegative than chlorine, so its nucleus attracts an incoming electron more strongly. However, the valence... | Chemistry 11th |
Which of the following will have the most negative electron gain enthalpy: P, S, Cl, F? Explain. | Electron gain enthalpy is the enthalpy change when a gaseous atom gains an electron. A more negative value indicates a greater affinity for the electron. Electron gain enthalpy generally becomes more negative across a period. Therefore, the expected order is $$Cl > S > P$$. Halogens (F, Cl) have the most negative value... | Chemistry 11th |
How does electronegativity vary (a) across a period and (b) down a group? Which element has the highest electronegativity? | Across a period (left to right), the effective nuclear charge increases and atomic radius decreases. This leads to a stronger attraction for bonding electrons, so electronegativity increases. Down a group (top to bottom), the number of electron shells increases. The outermost electrons are farther from the nucleus and ... | Chemistry 11th |
What is the Pauling scale of electronegativity? Why can electronegativity not be measured directly? | Electronegativity is the tendency of an atom to attract a shared pair of electrons in a chemical bond. The Pauling scale is a relative scale, ranging from approximately 0.7 to 3.98, that assigns a numerical value to an element's electronegativity. It is derived from bond-dissociation energies. Pauling's formula relates... | Chemistry 11th |
Why is chemical reactivity highest at the two extremes of a period and lowest in the centre? | Reactivity is an atom's tendency to gain, lose, or share electrons to achieve a stable, full outer electron shell (octet). Elements on the far left (Group 1) have only one valence electron. They can easily lose this electron, making them highly reactive metals. Elements on the far right (Group 17, Halogens) have seven ... | Chemistry 11th |
Explain the variation of metallic and non-metallic character (a) across a period and (b) down a group. | Across a period (left to right), the effective nuclear charge increases and atomic size decreases. This makes it harder for an atom to lose electrons, thus decreasing metallic character and increasing non-metallic character. Down a group (top to bottom), the number of electron shells increases, leading to a larger atom... | Chemistry 11th |
How does the reactivity of alkali metals compare with that of halogens? Explain in terms of ionization enthalpy and electron gain enthalpy. | Alkali metals (Group 1) have only one valence electron. They possess very low ionization enthalpies, meaning very little energy is needed to remove this electron to form a stable positive ion ($$M^+$$) with a noble gas configuration. This ease of losing an electron makes them extremely reactive. Halogens (Group 17) hav... | Chemistry 11th |
What are acidic, basic, and amphoteric oxides? How does the nature of oxides vary across a period? | Acidic oxides are generally oxides of non-metals. They react with water to produce an acid. For example, sulfur dioxide reacts with water to form sulfurous acid: $$SO_2 + H_2O \rightarrow H_2SO_3$$. Basic oxides are generally oxides of metals. They react with water to produce a base. For example, sodium oxide reacts wi... | Chemistry 11th |
What is shielding effect or screening effect? How does it influence the ionization enthalpy? | The shielding effect, or screening effect, describes the decrease in attraction between the nucleus and the valence electrons due to repulsion from the inner-shell electrons. This repulsion by inner electrons effectively reduces the positive nuclear charge experienced by the outer electrons. This reduced charge is call... | Chemistry 11th |
Why is the shielding effect more effective when inner shells are completely filled? How does this explain the low ionization enthalpy of alkali metals? | The shielding effect is the repulsion between inner-shell electrons and valence electrons, which reduces the net attractive force of the positive nucleus on the outer electrons. A completely filled inner shell, such as a noble gas configuration, has a dense and spherically symmetrical electron cloud. This stable arrang... | Chemistry 11th |
What are metalloids or semi-metals? Name four elements that are classified as metalloids. | Metalloids are chemical elements that have properties in between, or that are a mixture of, those of metals and nonmetals. Physically, they are often shiny (lustrous) like metals but are typically brittle solids and are only semi-conductors of electricity. On the periodic table, metalloids form a diagonal staircase-lik... | Chemistry 11th |
Why does metallic character increase down a group but decrease across a period? | Metallic character is the tendency of an atom to lose its valence electrons. Down a group, the principal energy level ($$n$$) increases, and shielding by inner electrons increases. This makes it easier to remove the outermost electrons, thus increasing metallic character. Across a period, the number of protons increase... | Chemistry 11th |
Explain the diagonal relationship between Be and Al in terms of their chemical properties. | The diagonal relationship between Beryllium ($$Be$$) and Aluminum ($$Al$$) arises because they have a similar charge-to-radius ratio ($$Z/r$$), leading to similar chemical behaviors. Both metals are rendered passive by the formation of a protective oxide layer ($$BeO$$ or $$Al_2O_3$$) on their surface, making them resi... | Chemistry 11th |
What is the total number of sigma and pi bonds in the following molecules?\
(a) $\mathrm{C}_{2}\mathrm{H}_{2}$\
(b) $\mathrm{C}_{2}\mathrm{H}_{4}$. | For ethyne ($$\mathrm{C}_{2}\mathrm{H}_{2}$$), the structure is H-C=C-H. A triple bond contains 1 sigma ($$\sigma$$) and 2 pi ($$\pi$$) bonds. Each C-H single bond is a sigma bond. Calculation for $$\mathrm{C}_{2}\mathrm{H}_{2}$$: Sigma bonds = (1 C-C) + (2 C-H) = 3. Pi bonds = 2. For ethene ($$\mathrm{C}_{2}\mathrm{H}... | Chemistry 11th |
Explain the structure of $\mathrm{CO}_{3}^{2-}$ ion in terms of resonance. | To draw the Lewis structure for the carbonate ion, carbon is the central atom bonded to three oxygen atoms. To satisfy the octet rule for all atoms, one C-O bond is a double bond and two are single bonds. The total valence electron count is $$4 + (3 \times 6) + 2 = 24$$. The position of the double bond is not fixed. It... | Chemistry 11th |
Explain the formation of a chemical bond. | Atoms are driven to bond to achieve a more stable electron configuration, typically by fulfilling the octet rule (having eight valence electrons). Bonding occurs through two primary mechanisms: the transfer of electrons from one atom to another (ionic bonding) or the sharing of electrons between atoms (covalent bonding... | Chemistry 11th |
Draw the Lewis structures for the following molecules and ions: $\mathrm{H}_{2}\mathrm{S}$, $\mathrm{SiCl}_{4}$, $\mathrm{BeF}_{2}$, $\mathrm{CO}_{3}^{2-}$, HCOOH. | For $\text{H}_2\text{S}$, the total valence electrons are $1(2) + 6 = 8$. Sulfur is the central atom, forming a single bond with each of the two hydrogen atoms. The remaining four electrons are placed on the sulfur atom as two lone pairs. For $\text{SiCl}_4$, the total valence electrons are $4 + 7(4) = 32$. Silicon is ... | Chemistry 11th |
Explain the important aspects of resonance with reference to the $\mathrm{CO}_{3}^{2-}$ ion. | The Lewis structure for the carbonate ion shows a central carbon atom bonded to three oxygen atoms. To satisfy the octet rule for all atoms, one carbon-oxygen bond is a double bond, and the other two are single bonds. The total valence electron count is 24. The position of the double bond is not fixed. It can be formed... | Chemistry 11th |
Define electronegativity. How does it differ from electron gain enthalpy? | Electronegativity is the qualitative tendency of an atom to attract a shared pair of electrons within a chemical bond. It is a dimensionless property. Electron gain enthalpy is the quantitative measure of the energy change ($$\Delta H$$) that occurs when a neutral atom in the gaseous phase gains an electron. The key di... | Chemistry 11th |
Apart from tetrahedral geometry, another possible geometry for $\mathrm{CH}_{4}$ is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why $\mathrm{CH}_{4}$ is not square planar? | The central carbon atom in $$\mathrm{CH}_{4}$$ has four valence electrons and forms four single covalent bonds with four hydrogen atoms. This results in four bonding pairs of electrons and zero lone pairs around the carbon. According to Valence Shell Electron Pair Repulsion (VSEPR) theory, these four electron pairs rep... | Chemistry 11th |
What is meant by hybridisation of atomic orbitals? Describe the shapes of $sp$, $sp^{2}$, $sp^{3}$ hybrid orbitals. | Hybridization is the process of mixing atomic orbitals of slightly different energies to form a new set of orbitals of equivalent energies and shapes. $sp$ hybridization involves mixing one s and one p orbital, resulting in two $sp$ hybrid orbitals. These orbitals are arranged linearly with a bond angle of $180^\circ$.... | Chemistry 11th |
What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type. | A bond pair consists of two electrons that are shared between two atoms, forming a covalent bond. These electrons are located in the region between the two nuclei. A lone pair, or non-bonding pair, is a pair of valence electrons that is not shared with another atom in a covalent bond. They belong exclusively to one ato... | Chemistry 11th |
Explain the formation of $\mathrm{H}_{2}$ molecule on the basis of valence bond theory. | A hydrogen atom (H) has one electron in its 1s atomic orbital. The electronic configuration is $$1s^1$$. According to Valence Bond Theory, a covalent bond forms from the overlap of two half-filled atomic orbitals. When two hydrogen atoms approach each other, their 1s orbitals overlap. The single electrons in each orbit... | Chemistry 11th |
Define hydrogen bond. Is it weaker or stronger than the van der Waals forces? | A hydrogen bond is an electrostatic attraction between a hydrogen atom covalently bonded to a highly electronegative atom (like nitrogen, oxygen, or fluorine) and another nearby electronegative atom. The bond is formed between the partially positive hydrogen ($$\delta+$$) and the lone pair of electrons on the partially... | Chemistry 11th |
Explain why molecules with incomplete octet exist. Give examples. | An incomplete octet occurs when the central atom in a molecule has fewer than eight electrons in its valence shell. This exception to the octet rule typically happens with central atoms that have few valence electrons, such as Beryllium (2) and Boron (3). For example, in Boron Trifluoride ($$\mathrm{BF_3}$$), the Boron... | Chemistry 11th |
What is a polar covalent bond? Explain with example. | A polar covalent bond is a type of chemical bond where a pair of electrons is unequally shared between two atoms. This occurs when there is a significant difference in the electronegativity of the two atoms. The more electronegative atom attracts the shared electrons more strongly. The unequal sharing creates partial c... | Chemistry 11th |
Explain why $\mathrm{H}_{2}\mathrm{O}$ has a net dipole moment while $\mathrm{BeF}_{2}$ has zero dipole moment. | Both molecules contain polar covalent bonds. In $$\mathrm{H}_{2}\mathrm{O}$$, oxygen is more electronegative than hydrogen. In $$\mathrm{BeF}_{2}$$, fluorine is more electronegative than beryllium. This creates a partial negative charge on the more electronegative atom and a partial positive charge on the less electron... | Chemistry 11th |
Compare the dipole moments of $\mathrm{NH}_{3}$ and $\mathrm{NF}_{3}$. Why is the dipole moment of $\mathrm{NH}_{3}$ greater? | Both $$\mathrm{NH}_{3}$$ and $$\mathrm{NF}_{3}$$ molecules have a trigonal pyramidal geometry with a lone pair of electrons on the central nitrogen atom. In $$\mathrm{NH}_{3}$$, nitrogen is more electronegative than hydrogen. The individual N-H bond dipoles point towards the nitrogen atom. The resultant of these bond d... | Chemistry 11th |
What is VSEPR theory? State its main postulates. | The shape of a molecule is determined by repulsions between all of the electron pairs present in the valence shell of the central atom. Electron pairs (both bonding and non-bonding/lone pairs) orient themselves in space to be as far apart as possible to minimize repulsion. Lone pairs of electrons occupy more space and ... | Chemistry 11th |
Why is a sigma bond stronger than a pi bond? | A sigma ($$\sigma$$) bond is formed by the direct, head-on overlap of atomic orbitals. The electron density is concentrated on the axis directly between the two bonded nuclei. A pi ($$\pi$$) bond is formed by the lateral, or sideways, overlap of parallel p-orbitals. The electron density is concentrated in two regions, ... | Chemistry 11th |
Explain $sp^{3}d^{2}$ hybridisation with the example of $\mathrm{SF}_{6}$. | The central atom is sulfur (S), with atomic number 16. Its ground-state valence electron configuration is $$3s^2 3p^4$$. It has only two unpaired electrons, suggesting it can form only two bonds. To form six bonds with fluorine, sulfur requires six unpaired electrons. An electron from the 3s orbital and an electron fro... | Chemistry 11th |
Describe the bonding in ethene ($\mathrm{C}_{2}\mathrm{H}_{4}$) molecule. | In ethene ($$\mathrm{C}_{2}\mathrm{H}_{4}$$), each carbon atom is $$sp^2$$ hybridized. They form three $$sp^2$$ hybrid orbitals and have one unhybridized p-orbital. A sigma ($$\sigma$$) bond is formed between the two carbon atoms by the head-on overlap of one $$sp^2$$ orbital from each carbon. Each carbon also forms tw... | Chemistry 11th |
Draw the energy level diagram for molecular orbitals of homonuclear diatomic molecules. | Atomic orbitals (AOs) from two identical atoms combine to form an equal number of molecular orbitals (MOs): lower-energy bonding MOs and higher-energy antibonding MOs. The s-orbitals overlap to form a sigma bonding ($$\sigma$$) and a sigma antibonding ($$\sigma^*$$) MO. The p-orbitals overlap to form one sigma ($$\sigm... | Chemistry 11th |
How is bond order calculated in molecular orbital theory? What does it indicate? | Bond order is defined as the number of chemical bonds between a pair of atoms. The formula to calculate bond order is: $$ \text{Bond Order} = \frac{(\text{Number of electrons in bonding MOs}) - (\text{Number of electrons in antibonding MOs})}{2} $$ A higher bond order value indicates greater bond stability and strength... | Chemistry 11th |
What is hydrogen bond? How is it formed? | A hydrogen bond is a special, strong type of dipole-dipole intermolecular force. It is not a covalent bond. It occurs when hydrogen ($$\text{H}$$) is covalently bonded to a highly electronegative atom like nitrogen ($$\text{N}$$), oxygen ($$\text{O}$$), or fluorine ($$\text{F}$$). The electronegative atom pulls the sha... | Chemistry 11th |
Describe the hybridisation in case of $\mathrm{PCl}_{5}$. Why are the axial bonds longer as compared to equatorial bonds? | The central atom in $\text{PCl}_5$ is phosphorus ($Z = 15$). Its ground state electronic configuration is $[\text{Ne}]3s^2 3p^3$. To form five bonds with chlorine, an electron from the 3s orbital is promoted to an empty 3d orbital. The excited state configuration becomes $[\text{Ne}]3s^1 3p^3 3d^1$. The one 3s, three 3... | Chemistry 11th |
Define bond angle. How does it help in determining the shape of a molecule? | The bond angle is the angle formed between three atoms across at least two bonds. It is typically measured in degrees. According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, electron pairs (both bonding and lone pairs) around a central atom repel each other. They arrange themselves to be as far apart as... | Chemistry 11th |
What is bond enthalpy? Why is average bond enthalpy used for polyatomic molecules? | Bond enthalpy is the amount of energy required to break one mole of a particular covalent bond in the gaseous state. This process is always endothermic, meaning it requires an energy input, so bond enthalpy values are positive. In polyatomic molecules (molecules with more than two atoms), the energy required to break a... | Chemistry 11th |
The skeletal structure of $\mathrm{CH}_{3}\mathrm{COOH}$ as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid. | First, calculate the total number of valence electrons for $\text{CH}_3\text{COOH}$. Carbon is in Group 14, Oxygen in Group 16, and Hydrogen in Group 1. Total valence electrons $= 2(\text{C}) + 4(\text{H}) + 2(\text{O}) = 2(4) + 4(1) + 2(6) = 8 + 4 + 12 = 24$. The octet rule states that atoms tend to bond in such a way... | Chemistry 11th |
How does the presence of lone pairs affect the bond angle in a molecule? Explain with examples. | According to VSEPR theory, electron pairs around a central atom repel each other and arrange themselves to be as far apart as possible. This arrangement determines the molecule's geometry. Lone pairs of electrons are held closer to the central atom's nucleus and are not shared with another atom. As a result, they occup... | Chemistry 11th |
Explain the structure of $\mathrm{NH}_{3}$ and $\mathrm{H}_{2}\mathrm{O}$ on the basis of hybridisation. | Structure of Ammonia ($$\mathrm{NH}_{3}$$): The central nitrogen atom (atomic number 7) has the valence shell electronic configuration $$2s^2 2p^3$$. To form three bonds with hydrogen, it undergoes $$sp^3$$ hybridisation. Four $$sp^3$$ hybrid orbitals are formed. Three of these orbitals overlap with the $$1s$$ orbitals... | Chemistry 11th |
Write the Lewis dot structure of CO molecule. | Find the total number of valence electrons. Carbon is in Group 14 and has 4 valence electrons. Oxygen is in Group 16 and has 6 valence electrons. Total valence electrons = $$4 + 6 = 10$$. Draw the skeletal structure by connecting the atoms with a single bond. This uses 2 electrons, leaving $$10 - 2 = 8$$ electrons. Dis... | Chemistry 11th |
Write the Lewis structure of the nitrite ion, $\mathrm{NO}_{2}^{-}$. | First, determine the total number of valence electrons. Nitrogen is in Group 15 (5 electrons), Oxygen is in Group 16 (6 electrons), and there is a -1 charge (1 electron). The formula for the total valence electrons is: (Valence e? of N) + 2 x (Valence e? of O) + (charge). Substituting the values gives: $$5 + 2(6) + 1 =... | Chemistry 11th |
Explain the structure of $\mathrm{CO}_{2}$ molecule. | The central atom is Carbon (C), and it is bonded to two Oxygen (O) atoms. Carbon has 4 valence electrons, and each Oxygen has 6, for a total of $$4 + 2(6) = 16$$ valence electrons. To satisfy the octet rule for all atoms, the Carbon atom forms two double bonds, one with each Oxygen atom. The resulting Lewis structure i... | Chemistry 11th |
Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br. | A Lewis dot symbol consists of the element's chemical symbol surrounded by dots that represent its valence electrons. Determine the number of valence electrons for each atom from its group in the periodic table. Na (Group 1) has 1. Mg (Group 2) has 2. B (Group 13) has 3. N (Group 15) has 5. O (Group 16) has 6. Br (Grou... | Chemistry 11th |
Write Lewis symbols for the following atoms and ions: S and $\mathrm{S}^{2-}$; Al and $\mathrm{Al}^{3+}$; H and $\mathrm{H}^{-}$. | First, determine the number of valence electrons for each neutral atom from its group on the periodic table. Sulfur (S) in Group 16 has 6 valence electrons. Aluminum (Al) in Group 13 has 3 valence electrons. Hydrogen (H) in Group 1 has 1 valence electron. For the neutral atoms, write the element symbol and place the co... | Chemistry 11th |
Define octet rule. Write its significance and limitations. | The octet rule states that atoms tend to gain, lose, or share electrons to achieve a full outer shell containing eight electrons, similar to that of a noble gas. Significance: The rule is fundamental for predicting the types of bonds atoms will form and the resulting molecular structures for many common compounds, espe... | Chemistry 11th |
Write the favourable factors for the formation of ionic bond. | Low ionization enthalpy of the metal atom. This allows for the easy formation of a positive ion (cation). High negative electron gain enthalpy of the non-metal atom. This facilitates the formation of a negative ion (anion). High lattice enthalpy of the ionic crystal. The large amount of energy released when ions form a... | Chemistry 11th |
Discuss the shape of the following molecules using the VSEPR model: $\mathrm{BeCl}_{2}$, $\mathrm{BCl}_{3}$, $\mathrm{SiCl}_{4}$, $\mathrm{AsF}_{5}$, $\mathrm{H}_{2}\mathrm{S}$, $\mathrm{PH}_{3}$. | For $\mathrm{BeCl}_{2}$, the central Be atom has 2 bonding pairs and 0 lone pairs, resulting in a linear shape. For $\mathrm{BCl}_{3}$, the central B atom has 3 bonding pairs and 0 lone pairs, resulting in a trigonal planar shape. For $\mathrm{SiCl}_{4}$, the central Si atom has 4 bonding pairs and 0 lone pairs, result... | Chemistry 11th |
Although geometries of $\mathrm{NH}_{3}$ and $\mathrm{H}_{2}\mathrm{O}$ molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss. | The central atom in $\text{NH}_3$ is N, which has 1 lone pair and 3 bond pairs. The central atom in $\text{H}_2\text{O}$ is O, which has 2 lone pairs and 2 bond pairs. According to VSEPR theory, the magnitude of repulsion between electron pairs follows the order: Lone Pair-Lone Pair (LP-LP) > Lone Pair-Bond Pair (LP-BP... | Chemistry 11th |
How do you express the bond strength in terms of bond order? | Bond order is the number of chemical bonds between a pair of atoms. A single bond has a bond order of 1, a double bond is 2, and a triple bond is 3. Bond strength is measured by bond dissociation energy, the energy required to break a bond. As the bond order increases, the number of shared electrons between atoms incre... | Chemistry 11th |
Define the bond length. | Bond length is defined as the equilibrium distance between the nuclei of two atoms joined by a chemical bond. It is an average distance because atoms vibrate, causing the internuclear distance to fluctuate. Bond length is influenced by factors like bond order and the size of the atoms. Higher bond order and smaller ato... | Chemistry 11th |
$\mathrm{H}_{3}\mathrm{PO}_{3}$ can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing $\mathrm{H}_{3}\mathrm{PO}_{3}$? If not, give reasons for the same. | The two structures provided are for the molecule $\mathrm{H}_{3}\mathrm{PO}_{3}$. Structure (1) shows one hydrogen atom directly bonded to the central phosphorus atom. Structure (2) shows all three hydrogen atoms bonded to oxygen atoms. The fundamental condition for resonance is that all canonical forms must have the s... | Chemistry 11th |
Write the resonance structures for $\mathrm{SO}_{3}$, $\mathrm{NO}_{2}$ and $\mathrm{NO}_{3}^{-}$. | For $$\mathrm{SO}_{3}$$, the central sulfur atom forms a double bond with one oxygen atom and single bonds with the other two. The position of the double bond can be interchanged among the three oxygen atoms, resulting in three equivalent resonance structures. For $$\mathrm{NO}_{2}$$, a radical, the central nitrogen at... | Chemistry 11th |
Use Lewis symbols to show electron transfer between the following atoms to form cations and anions:\
(a) K and S\
(b) Ca and O\
(c) Al and N. | For K and S: Potassium (K) has one valence electron, and sulfur (S) has six. Two potassium atoms each transfer one electron to a single sulfur atom. This forms two potassium cations ($$K^+$$) and one sulfide anion ($$S^{2-}$$). The resulting compound is $$K_2S$$. For Ca and O: Calcium (Ca) has two valence electrons, an... | Chemistry 11th |
Although both $\mathrm{CO}_{2}$ and $\mathrm{H}_{2}\mathrm{O}$ are triatomic molecules, the shape of $\mathrm{H}_{2}\mathrm{O}$ molecule is bent while that of $\mathrm{CO}_{2}$ is linear. Explain this on the basis of dipole moment. | In $$\mathrm{CO}_{2}$$, the two C=O bonds are polar. The molecule's linear geometry places these two bond dipoles at an angle of 180 to each other. The two bond dipoles are equal in magnitude and opposite in direction. They cancel each other out, resulting in a net dipole moment of zero for the molecule. In $$\mathrm{H... | Chemistry 11th |
Write the significance/applications of dipole moment. | To determine the polarity of a molecule. A molecule with a non-zero dipole moment is polar, while a molecule with a zero dipole moment is nonpolar. To help predict molecular geometry. For a molecule like $$\text{AB}_2$$, a zero dipole moment implies a linear structure (e.g., $$\text{CO}_2$$), while a non-zero dipole mo... | Chemistry 11th |
Explain with the help of suitable example polar covalent bond. | A polar covalent bond is a type of chemical bond where a pair of electrons is unequally shared between two atoms. This occurs when the atoms have different electronegativity values. Consider the Hydrogen Chloride ($$\text{HCl}$$) molecule. The electronegativity of Chlorine (3.16) is higher than that of Hydrogen (2.20).... | Chemistry 11th |
Arrange the bonds in order of increasing ionic character in the molecules: LiF, $\mathrm{K}_{2}\mathrm{O}$, $\mathrm{N}_{2}$, $\mathrm{SO}_{2}$ and $\mathrm{ClF}_{3}$. | The ionic character of a bond is determined by the difference in electronegativity ($$\Delta\text{EN}$$) between the two atoms. A larger difference indicates a more ionic bond. The formula for electronegativity difference is $$\Delta\text{EN} = |\text{EN}_A - \text{EN}_B|$$. The electronegativity values are: Li=0.98, F... | Chemistry 11th |
Explain why $\mathrm{BeH}_{2}$ molecule has a zero dipole moment although the Be–H bonds are polar. | The electronegativity difference between Beryllium (Be) and Hydrogen (H) creates a polar covalent bond. This results in a dipole for each Be-H bond. According to VSEPR theory, the central Be atom has two bonding electron pairs and no lone pairs. To minimize repulsion, these pairs are positioned 180 apart, giving the mo... | Chemistry 11th |
Which out of $\mathrm{NH}_{3}$ and $\mathrm{NF}_{3}$ has higher dipole moment and why? | Both $\mathrm{NH}_{3}$ and $\mathrm{NF}_{3}$ have a trigonal pyramidal shape with a lone pair of electrons on the central nitrogen atom. In $\mathrm{NH}_{3}$, the bond dipoles of the N-H bonds point towards the more electronegative nitrogen atom. The orbital dipole of the lone pair also points in the same direction. Th... | Chemistry 11th |
Describe the change in hybridisation (if any) of the Al atom in the following reaction: $\mathrm{AlCl}_{3} + \mathrm{Cl}^{-} \rightarrow \mathrm{AlCl}_{4}^{-}$. | In $$AlCl_3$$, the central Al atom is bonded to three Cl atoms and has no lone pairs. The steric number is 3, which corresponds to $$sp^2$$ hybridization and a trigonal planar geometry. In $$AlCl_4^-$$, a $$Cl^-$$ ion donates an electron pair to the central Al atom, forming a fourth bond. The Al atom is now bonded to f... | Chemistry 11th |
Is there any change in the hybridisation of B and N atoms as a result of the following reaction? $\mathrm{BF}_{3} + \mathrm{NH}_{3} \rightarrow \mathrm{F}_{3}\mathrm{B}.\mathrm{NH}_{3}$. | In the reactant $$BF_3$$, the Boron atom is bonded to three Fluorine atoms and has no lone pairs. In $$NH_3$$, the Nitrogen atom is bonded to three Hydrogen atoms and has one lone pair. Hybridization is determined by the sum of sigma bonds and lone pairs. For $$BF_3$$, Boron has 3 sigma bonds + 0 lone pairs = 3, which ... | Chemistry 11th |
Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in $\mathrm{C}_{2}\mathrm{H}_{4}$ and $\mathrm{C}_{2}\mathrm{H}_{2}$ molecules. | **Formation of Ethylene ($$\mathrm{C}_{2}\mathrm{H}_{4}$$) - Double Bond** Each carbon atom undergoes $$sp^2$$ hybridization, forming three $$sp^2$$ orbitals and leaving one unhybridized p-orbital. A sigma ($$\sigma$$) bond is formed between the two carbon atoms by the overlap of one $$sp^2$$ orbital from each. The rem... | Chemistry 11th |
Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?\
(a) $1s$ and $1s$\
(b) $1s$ and $2p_{x}$\
(c) $2p_{y}$ and $2p_{y}$\
(d) $1s$ and $2s$. | A sigma ($$\sigma$$) bond is formed by the head-on overlap of atomic orbitals along the internuclear axis. The x-axis is defined as this axis. S-orbitals are spherical. $$p_x$$ orbitals are oriented along the x-axis. $$p_y$$ orbitals are oriented along the y-axis, which is perpendicular to the internuclear axis. The ov... | Chemistry 11th |
Which hybrid orbitals are used by carbon atoms in the following molecules?\
(a) $\mathrm{CH}_{3}-\mathrm{CH}_{3}$\
(b) $\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}$\
(c) $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}$\
(d) $\mathrm{CH}_{3}-\mathrm{CHO}$\
(e) $\mathrm{CH}_{3}\mathrm{COOH}$. | To determine the hybridization of a carbon atom, count the number of sigma ($$\sigma$$) bonds it forms. Lone pairs are generally not present on carbon in stable organic molecules. If a carbon atom forms 4 sigma bonds (four single bonds), its hybridization is $$sp^3$$. If a carbon atom forms 3 sigma bonds (one double bo... | Chemistry 11th |
Distinguish between a sigma and a pi bond. | A sigma ($$\sigma$$) bond is formed by the direct, head-on (axial) overlap of atomic orbitals. A pi ($$\pi$$) bond is formed by the side-by-side (lateral) overlap of unhybridized p-orbitals. The electron density in a sigma bond is concentrated on the axis directly between the two bonded nuclei. In a pi bond, the electr... | Chemistry 11th |
Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals. | The combining atomic orbitals must have comparable energy. A significant energy difference prevents effective combination. The combining atomic orbitals must have the same symmetry about the molecular axis. For example, a $$p_x$$ orbital can combine with another $$p_x$$ orbital but not with a $$p_y$$ orbital. The combi... | Chemistry 11th |
Use molecular orbital theory to explain why the $\mathrm{Be}_{2}$ molecule does not exist. | A beryllium (Be) atom has the electron configuration $$1s^2 2s^2$$. The $$\mathrm{Be}_{2}$$ molecule, therefore, has a total of 8 electrons. The 8 electrons fill the molecular orbitals according to the Aufbau principle, resulting in the configuration: $$(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^... | Chemistry 11th |
Compare the relative stability of the following species and indicate their magnetic properties: $\mathrm{O}_{2}$, $\mathrm{O}_{2}^{+}$, $\mathrm{O}_{2}^{-}$ (superoxide), $\mathrm{O}_{2}^{2-}$ (peroxide). | The species given for comparison are $$\mathrm{O}_{2}$$ (16 electrons), $$\mathrm{O}_{2}^{+}$$ (15 electrons), $$\mathrm{O}_{2}^{-}$$ (17 electrons), and $$\mathrm{O}_{2}^{2-}$$ (18 electrons). Stability is determined by bond order, and magnetic properties are determined by the presence of unpaired electrons. The bond ... | Chemistry 11th |
Write the significance of a plus and a minus sign shown in representing the orbitals. | The sign refers to the phase of the electron's wave function, represented by the symbol $$\\psi$$. A plus (+) or minus (-) sign indicates the sign of the amplitude of the wave function in that region of space. It does not represent a positive or negative electrical charge. When orbitals combine to form bonds, same-phas... | Chemistry 11th |
What is meant by the term bond order? Calculate the bond order of: $\mathrm{N}_{2}$, $\mathrm{O}_{2}$, $\mathrm{O}_{2}^{+}$ and $\mathrm{O}_{2}^{-}$. | Bond order is half the difference between the number of electrons in bonding orbitals ($$N_b$$) and antibonding orbitals ($$N_a$$). The formula is $$Bond\ Order = \frac{1}{2}(N_b - N_a)$$. For $$\mathrm{N}_{2}$$ (14 electrons): $$N_b=10$$, $$N_a=4$$. Bond Order = $$\frac{1}{2}(10 - 4) = 3$$. For $$\mathrm{O}_{2}$$ (16 ... | Chemistry 11th |
What is a chemical bond? Explain why atoms combine to form molecules. | A chemical bond is an electrostatic force of attraction that holds atoms together in a molecule or compound. This attraction involves the valence (outermost) electrons of the atoms. Atoms form bonds to achieve a more stable electron configuration, similar to that of a noble gas. This principle is often guided by the oc... | Chemistry 11th |
Describe the Kössel-Lewis approach to chemical bonding. What is the basis of this approach? | The fundamental basis of this approach is the stability of noble gases, which have a complete octet of eight valence electrons ($ns^2np^6$), making them chemically inert. Atoms attempt to achieve this stable octet by either transferring or sharing their valence electrons. Kossel proposed that atoms can transfer electro... | Chemistry 11th |
What is Lewis symbol? How does it represent valence electrons? Give examples. | A Lewis symbol consists of the element's chemical symbol surrounded by dots. Each dot represents one valence electron, which is an electron in the outermost energy shell of an atom. The number of valence electrons is determined by the element's group in the periodic table. For example, Carbon is in Group 14 and has 4 v... | Chemistry 11th |
Explain the significance of Lewis symbols in determining the valence of an element. | A Lewis symbol represents an element with dots placed around its chemical symbol. Each dot signifies one valence electron. The number of dots in the Lewis symbol is equal to the number of valence electrons in the atom's outermost shell. Valence is the combining capacity of an element, determined by the number of electr... | Chemistry 11th |
What is an electrovalent or ionic bond? Explain its formation with the help of NaCl. | An electrovalent or ionic bond is formed by the complete transfer of one or more electrons from a metallic atom to a non-metallic atom. Sodium ($$\mathrm{Na}$$) has an electronic configuration of 2, 8, 1. To achieve a stable octet, it loses its single valence electron to become a sodium cation ($$\mathrm{Na}^{+}$$). Ch... | Chemistry 11th |
What is electrovalence? How is it related to the charge on an ion? | Electrovalence is defined as the number of electrons that an atom loses or gains to form an ionic bond. Atoms that lose electrons become positively charged ions (cations), and their electrovalence is positive. For example, a sodium atom loses one electron to form $$Na^+$$, so its electrovalence is +1. Atoms that gain e... | Chemistry 11th |
Define octet rule. Explain its significance in chemical bonding. | The octet rule states that atoms tend to bond in such a way that they each have eight electrons in their valence shells, giving them the same electronic configuration as a noble gas. Atoms can achieve an octet by either transferring electrons (ionic bonding) or sharing electrons (covalent bonding). For example, in sodi... | Chemistry 11th |
What are the limitations of the octet rule? Give examples. | The octet rule states that atoms bond to achieve eight valence electrons. However, there are three main exceptions to this rule. Incomplete Octet: The central atom has fewer than eight electrons. This is common for elements in Group 2 and 13. For example, in Boron Trifluoride ($$\mathrm{BF_3}$$), Boron only has six val... | Chemistry 11th |
What is a covalent bond? Explain with examples how covalent bonds are formed. | A covalent bond is a chemical link between two atoms where electron pairs are shared. This type of bond typically forms between non-metal atoms with similar electronegativity values. Atoms share electrons to achieve a more stable electron configuration, usually a full outer shell, as described by the octet rule. For ex... | Chemistry 11th |
Distinguish between single, double and triple covalent bonds with examples. | A single covalent bond is formed when two atoms share one pair of electrons. Example: In a methane molecule ($$\{CH_4\}$$), one carbon atom forms four single bonds with four hydrogen atoms. A double covalent bond is formed when two atoms share two pairs of electrons. Example: In an oxygen molecule ($$\{O_2\}$$), the tw... | Chemistry 11th |
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