p-Block Elements — Page 12 | JEE Chemistry

Q111

Given below are two statements. Statement I : Stannane is an example of a molecular hydride. Statement II : Stannane is a planar molecule. In the light of the above statement, choose the most appropriate answer from the options given below.

A Both Statement I and Statement II are true.

B Both Statement I and Statement II are false.

C Statement I is true but Statement II is false.

D Statement I is false but Statement II is true.

Correct Answer

Option C

Solution

Stannane or tin hydride is an inorganic compound with formula $\mathrm{SnH}_{4}$ $\therefore$ It is a non-planar molecule.

Q112

The correct order of bond enthalpy (

\mathrm{kJ~mol^{-1}}

) is :

A

\mathrm{C-C > Si-Si > Sn - Sn > Ge-Ge}

B

\mathrm{Si-Si > C-C > Ge-Ge > Sn-Sn}

C

\mathrm{C-C > Si-Si > Ge-Ge > Sn-Sn}

D

\mathrm{Si-Si > C-C > Sn-Sn > Ge-Ge}

Correct Answer

Option C

Solution

C–C bonds bond energy = 348 kJ mol-1 Si–Si bonds bond energy = 297 kJ mol-1 Ge–Ge bonds bond energy = 260 kJ mol-1 Sn–Sn bonds bond energy = 240 kJ mol-1 $\therefore$ Correct order is -

\mathrm{C-C > Si-Si > Ge-Ge > Sn-Sn}

Note : The property of forming bonds with atoms of the same element or tendency to self linking called catenation.

Carbon shows maximum catenation.

On moving down the group catenation tendency decreases.

This because the strength of $\mathrm{C}-\mathrm{C}$ bond is very high and in case of other elements, strength of ${M}-{M}$ (where ${M}={Si}, {G e}, {S n}, {P b}$ ) bond is decreases down the group.

Q113

The covalency and oxidation state respectively of boron in

\left[\mathrm{BF}_{4}\right]^{-}

, are :

A 3 and 4

B 4 and 3

C 4 and 4

D 3 and 5

Correct Answer

Option B

Solution

In the tetrafluoroborate anion (

\left[\mathrm{BF}_{4}\right]^{-}

), boron is bonded to four fluorine atoms through covalent bonds.

Therefore, the covalency of boron in this ion is 4.

The oxidation state of boron can be calculated by considering the charges on the atoms involved in the anion.

Fluorine has an oxidation state of -1, and there are four fluorine atoms in the anion, which contributes a total of -4.

Since the overall charge on the tetrafluoroborate anion is -1, the oxidation state of boron must be +3 in order to balance the charges.

So, the covalency and oxidation state of boron in

\left[\mathrm{BF}_{4}\right]^{-}

are 4 and 3, respectively.

Q114

For electron gain enthalpies of the elements denoted as

\Delta_{\mathrm{eg}} \mathrm{H}

, the incorrect option is :

A

\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{I})<\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{At})

B

\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{Te})<\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{Po})

C

\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{Cl})<\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{F})

D

\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{Se})<\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{S})

Correct Answer

Option D

Solution

First, recall that electron-gain enthalpy $\Delta_{\mathrm{eg}}H$ is often quoted as a (negative) numerical value.

A “more negative” electron-gain enthalpy means that an atom gains an electron more favorably.

1.

Known Trends in the Periodic Table Halogens (Group 17): Although one might expect the electron-gain enthalpy to become steadily less negative as we go down the group (due to increasing size and shielding), there is an anomaly at the top: $\text{Most negative: } \mathrm{Cl} > \mathrm{F} > \mathrm{Br} > \mathrm{I} > \mathrm{At} \quad(\text{in terms of magnitude of negativity}).$ Numerically (since these values are negative), that ordering translates to: $\Delta_{\mathrm{eg}}H(\mathrm{Cl}) The smaller (more negative) number on a number line corresponds to a larger (more exothermic) electron-gain enthalpy. Chalcogens (Group 16): A similar pattern holds, with an anomaly near oxygen. Going down from S to Se to Te to Po, the electron-gain enthalpy becomes less negative:$ \text{Most negative: } \mathrm{S} > \mathrm{Se} > \mathrm{Te} > \mathrm{Po} \quad(\text{in magnitude}).

$Numerically (all negative values), that means:$ \Delta_{\mathrm{eg}}H(\mathrm{S}) 2.

Checking Each Option We interpret each statement as a comparison of numerical values (remembering they are negative).

(A) $\,\Delta_{\mathrm{eg}}H(\mathrm{I}) Iodine’s electron-gain enthalpy is more negative than astatine’s. Numerically,$ \Delta_{\mathrm{eg}}H(\mathrm{I}) $is indeed a “smaller” (i.e. more negative) number than$ \Delta_{\mathrm{eg}}H(\mathrm{At}) $. So this statement is correct. (B)$ \,\Delta_{\mathrm{eg}}H(\mathrm{Te}) Tellurium (Te) is above polonium (Po) in Group 16, so Te typically has a more negative electron-gain enthalpy than Po.

Numerically, that means $\Delta_{\mathrm{eg}}H(\mathrm{Te})$ is smaller (more negative) than $\Delta_{\mathrm{eg}}H(\mathrm{Po})$ .

This statement is correct.

(C) $\,\Delta_{\mathrm{eg}}H(\mathrm{Cl}) Among halogens,$ \mathrm{Cl} $actually has the most negative$ \Delta_{\mathrm{eg}}H $. Fluorine’s value, while also negative, is a bit less so (because of small-orbital repulsion in F). Numerically, that means$ \Delta_{\mathrm{eg}}H(\mathrm{Cl}) $is indeed a smaller (more negative) number than$ \Delta_{\mathrm{eg}}H(\mathrm{F}) $. So this statement is correct. (D)$ \,\Delta_{\mathrm{eg}}H(\mathrm{Se}) In Group 16, sulfur (S) has a more negative electron-gain enthalpy than selenium (Se).

Numerically, that implies $\Delta_{\mathrm{eg}}H(\mathrm{S}) not the other way around. The statement says$ \Delta_{\mathrm{eg}}H(\mathrm{Se}) more negative than S—but that is not correct.

Hence, (D) is the incorrect statement.

Q115

The Lewis acid character of boron tri halides follows the order :

A

\mathrm{BI}_{3}>\mathrm{BBr}_{3}>\mathrm{BCl}_{3}>\mathrm{BF}_{3}

B

\mathrm{BCl}_{3}>\mathrm{BF}_{3}>\mathrm{BBr}_{3}>\mathrm{BI}_{3}

C

\mathrm{BF}_{3}>\mathrm{BCl}_{3}>\mathrm{BBr}_{3}>\mathrm{BI}_{3}

D

\mathrm{BBr}_{3}>\mathrm{BI}_{3}>\mathrm{BCl}_{3}>\mathrm{BF}_{3}

Correct Answer

Option A

Solution

Extent of back bonding, reduces down the group leading to more Lewis acidic strength.

Q116

Given below are two statements : Statement I: Upon heating a borax bead dipped in cupric sulphate in a luminous flame, the colour of the bead becomes green Statement II: The green colour observed is due to the formation of copper(I) metaborate In light of the above statements, choose the most appropriate answer from the options given below :

A Both Statement I and Statement II are false

B Both Statement I and Statement II are true

C Statement I is false but Statement II is true

D Statement I is true but Statement II is false

Correct Answer

Option A

Solution

On treatment with metal salt, boric anhydride forms metaborate of the metal which gives different colours in oxidising and reducing flame.

For example, in the case of copper sulphate, following reactions occur.

\mathrm{CuSO}_4+\mathrm{B}_2 \mathrm{O}_3 \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{\left( {Oxidation} \right)}^{Non - lu\min ous\,flame}} \mathrm{Cu}\left(\mathrm{BO}_2\right)_2+\mathrm{SO}_3

Two reactions may take place in reducing flame (Luminous flame) (i) The blue-green $\mathrm{Cu}\left(\mathrm{BO}_2\right)_2$ is reduced to colourless cuprous metaborate as :

\begin{aligned} & 2 \mathrm{Cu}\left(\mathrm{BO}_2\right)_2+2 \mathrm{NaBO}_2+\mathrm{C} \stackrel{\text{ Luminous }}{\longrightarrow} \\\\ & 2 \mathrm{CuBO}_2+\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+\mathrm{CO} \end{aligned}

(ii) Cupric metaborate may be reduced to metallic copper and bead appears red opaque.

\begin{aligned} & 2 \mathrm{Cu}\left(\mathrm{BO}_2\right)_2+4 \mathrm{NaBO}_2+2 \mathrm{C} \underset{\text{ flame }}{\stackrel{\text{ Luminous }}{\longrightarrow}} \\\\ & 2 \mathrm{Cu}+2 \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+2 \mathrm{CO} \end{aligned}

Q117

Given below are two statements: Statement I : Chlorine can easily combine with oxygen to form oxides; and the product has a tendency to explode. Statement II : Chemical reactivity of an element can be determined by its reaction with oxygen and halogens. In the light of the above statements, choose the correct answer from the options given below :

A Statement I is true but Statement II is false

B Both the Statements I and II are false

C Statement I is false but Statement II is true

D Both the Statements I and II are true

Correct Answer

Option D

Solution

Statement I is true because Chlorine (Cl) is a highly reactive halogen and has a strong tendency to form compounds with other elements, including oxygen.

When chlorine reacts with oxygen, it can form various oxides such as chlorine monoxide (ClO) and chlorine dioxide (ClO2).

These compounds can be explosive in certain conditions.

Statement II is also true because the chemical reactivity of an element can be determined by its ability to react with oxygen and halogens.

Elements that react readily with oxygen or halogens are considered to be highly reactive, while elements that do not react easily with these substances are considered to be less reactive.

The reactivity of an element can give important information about its behavior in chemical reactions and its potential uses in various industrial and technological applications.

Q118

During the borax bead test with

\mathrm{CuSO_4}

, a blue green colour of the bead was observed in oxidising flame due to the formation of :

A CuO

B Cu(BO

_2)_2

C Cu

_3

B

_2

D Cu

Correct Answer

Option B

Solution

Blue green colour is due to formation of $\mathrm{Cu}\left(\mathrm{BO}_{2}\right)_{2}$ $\mathrm{CuSO}_{4} \stackrel{\Delta}{\longrightarrow} \mathrm{CuO}+\mathrm{SO}_{3}$ $\mathrm{CuO}+\mathrm{B}_{2} \mathrm{O}_{3} \rightarrow \mathrm{Cu}\left(\mathrm{BO}_{2}\right)_{2}$

Q119

"A" obtained by Ostwald's method involving air oxidation of NH

_3

, upon further air oxidation produces "B", "B" on hydration forms an oxoacid of Nitrogen along with evolution of "A". The oxoacid also produces "A" and gives positive brown ring test. Identify A and B, respectively :

A

\mathrm{N_2O_3,NO_2}

B

\mathrm{NO_2,N_2O_5}

C

\mathrm{NO_2,N_2O_4}

D

\mathrm{NO,NO_2}

Correct Answer

Option D

Solution

Ostwald's process is : $4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \longrightarrow \underset{(A)}{4 \mathrm{NO}}+6 \mathrm{H}_{2} \mathrm{O}$ $\underset{\text{ (A) }}{2 \mathrm{NO}}+\mathrm{O}_{2} \longrightarrow \underset{\text{ (B) }}{2 \mathrm{NO}_{2}}$ $4 \mathrm{NO}_{2}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} \longrightarrow 4 \mathrm{HNO}_{3}$ $\therefore \mathrm{A}$ and $\mathrm{B}$ are $\mathrm{NO}$ and $\mathrm{NO}_{2}$ respectively

Q120

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R Assertion A : Carbon forms two important oxides - CO and CO

_2

. CO is neutral whereas CO

_2

is acidic in nature Reason R : CO

_2

can combine with water in a limited way to form carbonic acid, while CO is sparingly soluble in water In the light of the above statements, choose the most appropriate answer from the options given below :

A A is not correct but R is correct

B Both A and R are correct but R is NOT the correct explanation of A

C Both A and R are correct and R is the correct explanation of A

D A is correct but R is not correct

Correct Answer

Option C

Solution

Carbon monoxide is neutral and $\mathrm{CO}_{2}$ is acidic in nature because with the increase in oxidation state of carbon acidic strength increases.

So, Assertion is correct. $\mathrm{CO}_{2}$ combines with water to form carbonic acid while CO is sparingly soluble in water.

So, Reason is also correct and is the correct explanation of Assertion.