NEET PYQs

Hint

(d) Atomic and ionic radii of group 13 elements are lower than those of alkaline earth metals of group 2 primarily due to greater nuclear charge of group 13 elements as compared to group 2 elements. On moving down the group the atomic radius of Ga is slightly lower than that of AI. This is due to the presence of \(d\) – electrons in Ga which do not shield the nucleus effectively. As a result, the electrons in Ga experience greater force of attraction by the nucleus than in Al and hence the atomic radius of Ga 135 pm is slightly less than that of Al 143 pm. Thus, the increasing order of atomic radii of group 13 elements is \(\mathrm{B}<\mathrm{Ga}<\mathrm{Al}<\mathrm{In}<\mathrm{Tl}\)

Hint

(c) Boron does not have vacant \(d\)-orbitals in its valence shell, so it cannot expand its covalency beyond 4 i.e., B cannot form the ions like \(\mathrm{MF}_6^{3-}\).

Hint

(a) Inability of \(n s^2\) electrons of the valence shell to participate in bonding on moving down the group in heavier \(p\)-block elements is called inert pair effect. The inertness of \(s\)-subshell electrons towards bond formation is known as inert pair effect. This effect increases down the group thus, for \(\mathrm{Sn},+4\) oxidation state is more stable, whereas, for \(\mathrm{Pb},+2\) oxidation state is more stable, i.e. \(\mathrm{Sn}^{2+}\) is reducing while \(\)\mathrm{Pb}^{4+}[.latex] is, oxidising.

Hint

(d) The inertness of s-subshell electrons towards bond formation is known as inert pair effect. This effect increases down the group thus, for \(S n,+4\) oxidation state is more stable, whereas, for \(\mathrm{Pb},+2\) oxidation state is more stable, i.e., \(\mathrm{Sn}^{2+}\) is reducing while \(\mathrm{Pb}^{4+}\) is oxidizing.

Hint

(b) \(\mathrm{AlF}_3\) is insoluble in anhydrous HF because the \(\mathrm{F}^{-}\)ions are not available in hydrogen bonded HF molecules but, it becomes soluble in presence of little amount of KF due to formation of complex, \(\mathrm{K}_3\left[\mathrm{AlF}_6\right]\). \(\mathrm{AlF}_3+3 \mathrm{KF} \rightarrow \mathrm{K}_3\left[\mathrm{AlF}_6\right]\)

Hint

(d) Isostructural species are those which have the same shape as well as same hybridisation.
In option (a), they both have the same hybridisation i.e., \(\mathrm{sp}^3\) as well as the same shape i.e., pyramidal. Hence, these are isostructural.
In option (b), \(\mathrm{XeF}_4\) has a square planar shape and \(\mathrm{XeO}_4\) is tetrahedral. So, they are not isostructural.

In option (c), both have the same hybridisation i.e., \(\mathrm{sp}^3\) and the same structure i.e., tetrahedral. Hence, these are isostructural.
In option (d), both are giant covalent structures with the same hybridisation i.e., \(\mathrm{sp}^3\) and tetrahedral geometry. Hence, these are isostructural.

Hint

(c) Boric acid behaves as a Lewis acid, by accepting a pair of electrons from \(\mathrm{OH}^{-}\)ion of water thereby releasing a proton.

Hint

(b) In group 13 elements, stability of +3 oxidation state decreases down the group while that of +1 oxidation state increases due to inert pair effect. Hence, stability of +1 oxidation state increases in the sequence: \(\mathrm{Al}<\mathrm{Ga}<\mathrm{In}<\mathrm{Tl}\).

Hint

(d) BN is known as inorganic graphite and has structure similar to graphite.

Hint

(b) Boron trifluoride \(\mathrm{BF}_3\) is least likely to behave as Lewis base.
On the contrary, it is a Lewis acid.
In \(\mathrm{BF}_3\), the central B atom has six electrons and can easily accept an electron pair from a suitable base.
Thus, it acts as Lewis acid. It cannot act as Lewis base as it cannot donate an electron pair.

Hint

(b) \(\mathrm{SnO}_2\) :
– Amphoteric oxides are metal oxides that react with acids and bases to form salts and water.
– Tin oxide \(\left(\mathrm{SnO}_2\right)\) is an amphoteric in nature because it can react with both acid and base to form salts and water.
– The reaction of Tin oxide with hydrochloric acid \((\mathrm{HCl})\) is as follows:
\(
\mathrm{SnO}_2+4 \mathrm{HCl} \rightarrow \mathrm{Cl}_2+\mathrm{SnCl}_2+2 \mathrm{H}_2 \mathrm{O}
\)
– The reaction of Tin oxide with potassium hydroxide \((\mathrm{KOH})\) is as follows:
\(
\mathrm{SnO}_2+2 \mathrm{KOH} \rightarrow \mathrm{~K}_2 \mathrm{SnO}_3+\mathrm{H}_2 \mathrm{O}
\)
The explanation for the incorrect option:
(a) CaO :
Calcium oxide \((\mathrm{CaO})\) is basic in nature, not amphoteric.
(b) \(\mathrm{CO}_2\) :
Carbon dioxide \(\left(\mathrm{CO}_2\right)\) is acidic in nature, not amphoteric.
(c) \(\mathrm{SiO}_2\) :
Silicon dioxide \(\left(\mathrm{SiO}_2\right)\) is acidic in nature, not amphoteric.
Therefore, the correct option is (b). \(\mathrm{SnO}_2\) is amphoteric in nature.

Hint

(d) Aluminium dissolve in excess NaOH to liberating hydrogen and forming metaaluminate

\(
2 \mathrm{Al}+2 \mathrm{NaOH}+6 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right]
\)
\(
\text { or }\left(2 \mathrm{NaAlO}_2 \cdot 2 \mathrm{H}_2 \mathrm{O}\right)+3 \mathrm{H}_2
\)

Hint

(c) \(\mathrm{Al}_2 \mathrm{O}_3\) may be converted into anhyd. \(\mathrm{AlCl}_3\) by heating a mixture of \(\mathrm{Al}_2 \mathrm{O}_3\) and carbon in dry chlorine.
\(\mathrm{Al}_2 \mathrm{O}_3+3 \mathrm{C}+3 \mathrm{Cl}_2 \longrightarrow \mathrm{Al}_2 \mathrm{Cl}_6 ~22 \mathrm{~mm}\) Hot and dry ~8 mm Anhy \(\mathrm{AlCl}_2\)

Hint

(a) \(\mathrm{SiCl}_4\) gets hydrolysed in moist air and gives white fumes which are used as a smoke screen in warfare.

Hint

(a) Carbon has no \(d\)-orbitals, while silicon contains \(d\)-orbitals in its valence shell which can be used for bonding purposes.

Hint

(a) When \(n s^2\) electrons of outermost shell do not participate in bonding then these \(n s^2\) electrons are called inert pair and the effect is called inert pair effect. Due to this inert pair effect \(\mathrm{Ge}, \mathrm{Sn}\) and Pb of group 14 have a tendency to form both +4 and +2 ions. Now the inert pair effect increases down the group, hence the stability of \(M^{2+}\) ions increases and \(M^{4+}\) ions decreases down the group. For this reason, \(\mathrm{Pb}^{2+}\) is more stable than \(\mathrm{Pb}^{4+}\) and \(\mathrm{Sn}^{4+}\) is more stable than \(\mathrm{Sn}^{2+}\).

Hint

(a) Except diamond other three conduct electricity. Potassium and sodium are metallic conductors, while graphite is a non-metallic conductor.

Hint

(c) Lead pencil contains graphite and clay. It does not contain lead.

Hint

(c) In graphite each carbon atom undergoes \(s p^2\)-hybridisation and is covalently bonded to three other carbon atoms by single bonds. The fourth electron forms \(\pi\)-bond. The electrons are delocalised over the whole sheet i.e. electrons are spread out between the structure.

Hint

(b) In diamond, each carbon atom is \(s p^3\) hybridized and thus, forms covalent bonds with four other carbon atoms lying at the corners of a regular tetrahedron.

Hint

(c) All the above are conductors except diamond. Diamond is an insulator.

Hint

(d) The incorrect statement is: “Common oxidation states of group 14 elements are +1 and +3. The correct statement is that the common oxidation states of group 14 elements are +2 and +4 .

Explanation: Group 14 elements typically exhibit oxidation states of +2 and +4 due to the “inert pair effect” where the heavier elements in the group tend to show a more stable +2 oxidation state.

Key points about group 14 oxidation states:
– Elements in group 14: Carbon (C), Silicon (Si), Germanium (Ge), Tin (Sn), and Lead (Pb)
– Common oxidation states: +2 and +4
– Inert pair effect: Explains why heavier elements like tin and lead favor the +2 oxidation state more than the +4 state.

Hint

(c) is the incorrect statement.
– The stability of the +1 oxidation state increases down the group 13 due to the inert pair effect
– Boron trioxide is acidic, aluminium and gallium tri oxides are amphoteric, and indium and thallium tri oxides are basic
– The hybridization of Al in \(\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}\) is \(\mathrm{sp}^3 \mathrm{~d}^2\)
– Boron has two isotopes, \({ }^{10} \mathrm{~B}\) and \({ }^{11} \mathrm{~B}\)
(a) The statement is correct. The stability of the +1 oxidation state increases down the group 13 due to the inert pair effect.
(b) The statement is correct. Boron trioxide is acidic, aluminium and gallium tri oxides are amphoteric, and indium and thallium tri oxides are basic.
(c)
\(
\text { The statement is incorrect. The hybridization of } \mathrm{Al} \text { in }\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \text { is } \mathrm{sp}^3 \mathrm{~d}^2 \text {, not } \mathrm{d}^2 \mathrm{sp}^3 \text {. }
\)
(d)
\(
\text { The statement is correct. Boron has two isotopes, }{ }^{10} \mathrm{~B} \text { and }{ }^{11} \mathrm{~B} \text {. }
\)

Hint

(a) Option a: The acidic strength of \(\mathrm{HX}(\mathrm{X}=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}\) and I\()\) follows the order: \(\mathrm{HF}>\mathrm{HCl}>\mathrm{HBr}>\mathrm{HI}\).

This statement is incorrect. The correct order of acidic strength for hydrohalic acids is \(\mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{HI}\). This is because as we move down the group, the bond strength between hydrogen and the halide decreases, making it easier to ionize in water and making the acid stronger.

Option b: Fluorine exhibits -1 oxidation state whereas other halogens exhibit \(+1,+3,+5\) and +7 oxidation states also.

This statement is correct. Fluorine is the most electronegative element and can only exhibit -1 oxidation state. Other halogens can exhibit positive oxidation states due to the availability of d-orbitals.

Option c: The enthalpy of dissociation of \(\mathrm{F}_2\) is smaller than that of \(\mathrm{Cl}_2\).
This statement is correct. The bond dissociation energy of \(\mathrm{F}_2\) is indeed smaller than that of \(\mathrm{Cl}_2\) because the \(\mathrm{F}-\mathrm{F}\) bond is weaker due to significant electron-electron repulsion between the non-bonding electrons in the small fluorine molecule.

Option d: Fluorine is a stronger oxidising agent than chlorine.
This statement is correct. Fluorine is the most electronegative element, which makes it the strongest oxidizing agent among the halogens.

Therefore, the incorrect statement is Option a.

Hint

(1)

Hint

(d) The elements in Group 16 of the periodic table, also known as the chalcogens, include oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and polonium (Po). Generally, these elements are known to commonly exhibit a -2 oxidation state because they have six electrons in their outermost shell and can gain two electrons to complete their octet, thereby achieving a stable electronic configuration similar to the nearest noble gas.

Oxygen, being the most electronegative element in this group, predominantly shows a -2 oxidation state in most of its compounds, such as water \(\left(\mathrm{H}_2 \mathrm{O}\right)\) and oxides like \(\mathrm{SO}_2\).

Selenium and Tellurium also typically exhibit the -2 oxidation state. For instance, in compounds like \(\mathrm{H}_2 \mathrm{Se}\) (hydrogen selenide) and \(\mathrm{H}_2 \mathrm{Te}\) (hydrogen telluride), these elements are in the -2 state.

Polonium (Po), on the other hand, is different from the lighter chalcogens. Being a metalloid with more metallic character, it does not favor the -2 oxidation state nearly as strongly as the other elements in its group. Polonium most commonly exhibits +2 and +4 oxidation states in its compounds, like \(\mathrm{PoO}_2\) and \(\mathrm{PoCl}_2\). Due to the relativistic effects and its position in the periodic table, the -2 oxidation state is unstable and rare in polonium compounds.

Given this, the correct answer is Option d: Po. Polonium does not commonly exhibit the -2 oxidation state like the other Group 16 elements.

Hint

(b) Coke : It is used as reducing agent in carbon reduction methods. (in metallurgical process)
Diamond: It is a allotrope of carbon in which each carbon is \(\mathrm{sp}^3\) hybridised.
Fullerene : It contains pentagonal & hexagonal rings (cage like structure)
Graphite : It is soft solid because graphite layers are bonded with weak Vander Wall attractions.

Hint

(a) due to Inert pair effect in \(\mathrm{TlI}\) (+1) Oxidation State is more stable than (+3) Oxidation State in \(\mathrm{TlI}_3\).

Hint

(d) Maximum covalency of boron is four. Boron cannot form \(\left[\mathrm{M}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\) due to absence of d-orbital in boron.

Hint

(b) 

(a) \(\mathrm{Al}^{+} \text {is unstable in solution because } \mathrm{SRP} \text { of } \mathrm{Al}^{+} / \mathrm{Al} \text { is positive hence, it will get reduced in solution. }\)
(b) \(
\mathrm{Tl} / \mathrm{Tl}^{+}=+0.34 \text { (SOP) }
\)
\(
\mathrm{Tl} / \mathrm{Tl}^{3+}=-1.26 \text { (SOP) }
\)
Since, SOP \(\mathrm{Tl} / \mathrm{Tl}^{+}>\mathrm{Tl} / \mathrm{Tl}^{3+}\)
Hence, Tl will get oxidised to \(\mathrm{Tl}^{+}\)easily than \(\mathrm{Tl}^{3+}\)
(c) \(\text { Since } \mathrm{Al} / \mathrm{Al}{ }^{3+} \text { has more SOP than } \mathrm{Tl} / \mathrm{Tl}^{+} \text {it will be more electropositive }\)
(d) \(\text { Since } \mathrm{Tl}^{3+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Tl}^{+} ; \mathrm{E}=\frac{3 \times 1.26+0.34}{2}\)
\(
\begin{aligned}
&=+2.04\\
&\text { Hence, } \mathrm{Tl}^{3+} \text { is better oxidising agent not reducing agent }
\end{aligned}
\)

Hint

(a)

Diborane is a type of boron hydride, which is a compound of boron and hydrogen.The chemical formula of diborane is \(\mathbf{B}_2 \mathrm{H}_6\), i.e., two boron atoms and six hydrogen atoms, hence it is also known as diboron hexahydride. The hybridization of the boron atom in diborane is \(\mathbf{s p}^3\) and has four hybrid orbitals. The structure of diborane is so that the four terminal hydrogen atoms and the two boron atoms lie in a single plane. There are two types of bonding in diborane, 2-center-2-electron bonds, and 3-center-2-electron bonds. The four terminal B-H bonds have a 2-center-2-electron bond and the central \(\mathrm{B}-\mathrm{H}-\mathrm{B}\) have a 3-center-2-electron bond. The 3-center-2-electron bond is also called a banana bond.Hence, from the above explanation, the incorrect statement about diborane is 1) Both the Boron atoms are \(\mathbf{s p}^{\mathbf{2}}\) hybridized.

Hint

\(
\text { Option } d \text { : Diamond is } \mathrm{sp}^3 \text { hybridised and graphite is } \mathrm{sp}^2 \text { hybridized. }
\)

Concept:
Allotropes of Carbon – Carbon can exist in more than one physical state due to its catenation property. The allotropes of carbon are
1. Diamond – It is 3-D crystalline in nature, carbon atoms linked tetrahedrally in the diamond.
– The hybridization of each carbon atom is \(\mathrm{sp}^3\).
– It is the hardest form of carbon and has a high melting point.
– It is a poor conductor of electricity.
2. Graphite – Graphite is a two-dimensional flat form of carbon. It is a honeycomblike structure having multilayer Graphene sheets stacked on one another.
– In its structure, each carbon atom is linked to three other carbon atoms through covalent bonds. The fourth bond is the \(\pi\) bond.
– The hybridization of each carbon atom is \(\mathrm{sp}^2\).
– The electron in the pi bond participates in the conduction of electricity.
3. Fullerene – It consists at least 60 carbon atoms.
– Carbon atoms joined to form closed or partially closed cage-like structures often called buckyballs.
– Fullerene has a different structure at different pressures.

Explanation:
The correct statement about diamond and graphite is “Diamond is \(\mathbf{s p}^3\) hybridized and graphite is sp \({ }^2\) hybridized”.
As in diamond –
– 3-D structures of carbon atoms joined together by covalent bonds are found.
– Each carbon atom is tetrahedrally connected to other carbon atoms with a covalent bond.
– Thus, the hybridization of each carbon atom in diamond is \(\mathbf{s p}^3\).
In the case of graphite –
– It is a 2-D structure of carbon atoms joined together in form of layers of graphene.
– each carbon atom is connected to three other carbon atoms, by \(3 \sigma\) and \(1 \pi\) bonds.
– Thus, the hybridization of each carbon atom in graphite is \(\mathbf{s p}^2\).
Other statements are incorrect as-
– Diamond cannot be used as a dry lubricant because of its extreme hardness.
– Both diamond and graphite are covalent solid.
– Diamond is a 3-D structure and graphite is 2-D.
Hence the correct answer is option d.

Hint

(c) The stability of the +1 oxidation state increases in the following order: \(\mathrm{Al}^+\) < \(\mathrm{Ga}^{+}\) < \(\mathrm{In}^{+}\) < \(\mathrm{Tl}^{+}\).

Explanation: As you move down the group in the periodic table, the stability of the +1 oxidation state increases due to the “inert pair effect”. Therefore, \(\ln +\) is more stable than \(\mathrm{Ga}+\) and \(\mathrm{Al}+\) is more stable than \(\mathrm{Ga}+\).

Hint

(b) – Borax: \(\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O}=\mathrm{Na}_2\left[\mathrm{~B}_4 \mathrm{O}_5(\mathrm{OH})_4\right] \cdot 8 \mathrm{H}_2 \mathrm{O}\)
– Kernite : \(\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 4 \mathrm{H}_2 \mathrm{O}\)
– Orthoboric acid: \(\mathrm{H}_3 \mathrm{BO}_3=\mathrm{B}(\mathrm{OH})_3\)
– Borax bead : \(\mathrm{NaBO}_2\) (Sodium metaborate)

Hint

(c)

\(
\begin{aligned}
&\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \xrightarrow{\Delta} \mathrm{~B}_2 \mathrm{O}_3+2 \mathrm{NaBO}_2\\
&\text { Product } \mathrm{X} \text { is } \mathrm{B}_2 \mathrm{O}_3
\end{aligned}
\)

Hint

(c) Dry ice is used as refrigerant. \(\mathrm{C}_{60}\) contains 20 six membered ring, 12 five membered rings. In the presence of a zeolite catalyst (HZSM-5), ethanol, methanol and larger alcohols can be converted into gasoline at \(300-400^{\circ} \mathrm{C} . \mathrm{CO}\), is a tasteless, colorless, and odorless gas.

Hint

(a)

Among the given options, \(\mathrm{SnO}_2\) (tin oxide) is amphoteric in nature, meaning it can react with both acids and bases to form salts and water.

Explanation:
– \(\mathrm{SnO}_2\): This oxide can react with acids like hydrochloric acid and bases like potassium hydroxide, making it amphoteric.
– \(\mathrm{SIO_2}\) (silicon dioxide), \(\mathrm{GeO}_2\) (germanium dioxide), and \(\mathrm{CO}_2\) (carbon dioxide): These are all considered acidic oxides.

Key point: An amphoteric oxide is a metal oxide that can act as both an acid and a base depending on the reaction.

Hint

(b)

\(
\mathrm{PbF}_4 \text { is covalent in nature }
\)
Explanation:
– \(\mathrm{PbF}_4\) is ionic in nature according to Fajan’s rule as Pb has a large size as compared to F. On the other hand, there is more electronegativity difference between F and Pb which make the bond ionic.
– \(\mathrm{SiCl}_4\) is easily hydrolyzed as Si has empty d orbitals which makes a feasible hydrolysis reaction. The lone pair of water can easily occupy the empty d orbitals.
– Ge forms stable complex in +4 oxidation state as compared to in +2 oxidation state. Because in +4 oxidation state Ge acquires exact full filled configuration and is a stable one.
– \(\mathrm{SnF}_4\) is ionic in nature based on Fajan’s rule. There is a large difference in size as well as electronegativities of the atoms which make the bond ionic between Sn and F .

So, based on the above explanation the first statement ‘ \(\mathrm{PbF}_4\) is covalent in nature’ is incorrect as it is an ionic molecule.

Hint

(a)

The species that is not stable is \(\left[\mathrm{SiCl}_6\right]^{2-}\).
Explanation: While elements like Si, Ge, and Sn can form stable octahedral complexes with smaller anions like fluoride (e.g., \(\left[\mathrm{SiCl}_6\right]^{2-}\) ) due to the availability of d-orbitals, the large size of chloride ions causes significant steric hindrance when trying to surround a small silicon atom, making \(\left[\mathrm{SiCl}_6\right]^{2-}\) unstable.

Key points:
Steric hindrance:
The large size of chloride ions creates repulsion when trying to fit six of them around the smaller silicon atom.
d-orbital involvement:
The ability of elements like \(\mathrm{Si}, \mathrm{Ge}\), and Sn to form stable octahedral complexes is due to the presence of d-orbitals which can accommodate additional ligands.

Hint

(a) Write the equation for the hydrolysis of \(A l^{3+}\) ions, then determine the formula of the complex ion and the hybridization state of \(A l\).
\(
\begin{aligned}
&\text { Write the equation for the hydrolysis of } A l^{3+} \text { ions. }\\
&\mathrm{Al}^{3+}+6 \mathrm{H}_2 \mathrm{O} \rightarrow\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}+3 \mathrm{H}^{+}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Determine the formula of the complex ion. }\\
&\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}
\end{aligned}
\)
Determine the hybridization state of \(\boldsymbol{A l}\).

The coordination number of \(A l\) in the complex ion is 6 . The hybridization state of \(A l\) is \(s p^3 d^2\).
The formula of the complex A is \(\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}\) and the hybridization state of Al in \(A\) is \(s p^3 d^2\).