Match List-I with List-II : List-I (Species) List-II (Hybrid Orbitals) (a) SF_4 (i) sp^3d^2 (b) IF_5 (ii) d^2sp^3 (c) NO_2^ + (iii) sp^3d (d) NH_4^ + (iv) sp^3 (v) sp Choose the correct answer from the options given below :

(a)-(iii), (b)-(i), (c)-(v) and (d)-(iv)

(a)-(i), (b)-(ii), (c)-(v) and (d)-(iii)

(a)-(ii), (b)-(i), (c)-(iv) and (d)-(v)

(a)-(iv), (b)-(iii), (c)-(ii) and (d)-(v)

Chemical Bonding and Molecular Structure PYQ — JEE Main 2021 · 22 Jul · Shift 2 · Q7 with Solution

Match List-I with List-II :

	List-I (Species)		List-II (Hybrid Orbitals)
(a)	$S{F_4}$	(i)	$s{p^3}{d^2}$
(b)	$I{F_5}$	(ii)	${d^2}s{p^3}$
(c)	$NO_2^ +$	(iii)	$s{p^3}d$
(d)	$NH_4^ +$	(iv)	$s{p^3}$
		(v)	$sp$

Choose the correct answer from the options given below :

A
(a)-(i), (b)-(ii), (c)-(v) and (d)-(iii)
B
(a)-(ii), (b)-(i), (c)-(iv) and (d)-(v)
C
(a)-(iii), (b)-(i), (c)-(v) and (d)-(iv)
D
(a)-(iv), (b)-(iii), (c)-(ii) and (d)-(v)

View written solutionFree

Correct answer: C

Determine the hybridization of each species in List-I

We use steric number = number of bonded atoms + number of lone pairs on the central atom.

For $(a)\; SF_4$

Central atom: $S$
Sulfur has 6 valence electrons.
It forms 4 bonds with $F$ and has 1 lone pair.
Thus, total electron pairs around $S$ = $5$ .

So, hybridization is: $sp^3d$ Hence, $(a) \to (iii)$

For $(b)\; IF_5$

Central atom: $I$
Iodine has 7 valence electrons.
It forms 5 bonds with $F$ and has 1 lone pair.
Total electron pairs around $I$ = $6$ .

So, hybridization is: $sp^3d^2$ Hence, $(b) \to (i)$

For $(c)\; NO_2^+$

Central atom: $N$
Structure is linear: $O=N=O$
Around nitrogen, there are 2 regions of electron density and no lone pair.

So, hybridization is: $sp$ Hence, $(c) \to (v)$

For $(d)\; NH_4^+$

Central atom: $N$
Nitrogen forms 4 bonds and has no lone pair in ammonium ion.
Total electron pairs around $N$ = $4$ .

So, hybridization is: $sp^3$ Hence, $(d) \to (iv)$

Final matching

Thus the correct matching is: $(a)-(iii),\quad (b)-(i),\quad (c)-(v),\quad (d)-(iv)$

This corresponds to Option C.

Comparison with stored answer

Stored correct answer = C

Our derived answer = C

So, they agree.

More from Chemical Bonding and Molecular Structure

Match List-I with List-II :

	List-I (Species)		List-II (Hybrid Orbitals)
(a)	$S{F_4}$	(i)	$s{p^3}{d^2}$
(b)	$I{F_5}$	(ii)	${d^2}s{p^3}$
(c)	$NO_2^ +$	(iii)	$s{p^3}d$
(d)	$NH_4^ +$	(iv)	$s{p^3}$
		(v)	$sp$

Choose the correct answer from the options given below :

A
(a)-(i), (b)-(ii), (c)-(v) and (d)-(iii)
B
(a)-(ii), (b)-(i), (c)-(iv) and (d)-(v)
C
(a)-(iii), (b)-(i), (c)-(v) and (d)-(iv)
D
(a)-(iv), (b)-(iii), (c)-(ii) and (d)-(v)

View written solutionFree

Correct answer: C

Determine the hybridization of each species in List-I

We use steric number = number of bonded atoms + number of lone pairs on the central atom.

For $(a)\; SF_4$

Central atom: $S$
Sulfur has 6 valence electrons.
It forms 4 bonds with $F$ and has 1 lone pair.
Thus, total electron pairs around $S$ = $5$ .

So, hybridization is: $sp^3d$ Hence, $(a) \to (iii)$

For $(b)\; IF_5$

Central atom: $I$
Iodine has 7 valence electrons.
It forms 5 bonds with $F$ and has 1 lone pair.
Total electron pairs around $I$ = $6$ .

So, hybridization is: $sp^3d^2$ Hence, $(b) \to (i)$

For $(c)\; NO_2^+$

Central atom: $N$
Structure is linear: $O=N=O$
Around nitrogen, there are 2 regions of electron density and no lone pair.

So, hybridization is: $sp$ Hence, $(c) \to (v)$

For $(d)\; NH_4^+$

Central atom: $N$
Nitrogen forms 4 bonds and has no lone pair in ammonium ion.
Total electron pairs around $N$ = $4$ .

So, hybridization is: $sp^3$ Hence, $(d) \to (iv)$

Final matching

Thus the correct matching is: $(a)-(iii),\quad (b)-(i),\quad (c)-(v),\quad (d)-(iv)$

This corresponds to Option C.

Comparison with stored answer

Stored correct answer = C

Our derived answer = C

So, they agree.

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