Match List-I with List-II

List I (Species) List II (Electronic distribution)

A. ${Cr}^{+ 2}$ i. $3{d}^{8}$

B. ${Mn}^{+}$ ii. $3{d}^{3}4{s}^{1}$

C. ${Ni}^{+ 2}$ iii. . $3{d}^{4}$

D. ${V}^{+}$ iv. $3{d}^{5}4{s}^{1}$

Choose the correct answer from the options given below:

Question

Match List-I with List-II

	List I (Species)		List II (Electronic distribution)
A.	${Cr}^{+ 2}$	i.	$3{d}^{8}$
B.	${Mn}^{+}$	ii.	$3{d}^{3}4{s}^{1}$
C.	${Ni}^{+ 2}$	iii.	. $3{d}^{4}$
D.	${V}^{+}$	iv.	$3{d}^{5}4{s}^{1}$

Choose the correct answer from the options given below:

Quantrex Academy · Accepted Answer

Correct answer: B. (A)-III, (B) - IV, (C) - I, (D)-II

Electrons must be removed from the outermost orbital to form an ion.

The electronic configurations of the elements are as follows,

$Cr 24 \to [ Ar ] 3{d}^{5}4{s}^{1}; {Cr}^{2 +} \to [ Ar ] 3{d}^{4}$

$Mn 25 \to [ Ar ] 3{d}^{5}4{s}^{2}; {Mn}^{+} \to [ Ar ] 3{d}^{5}4{s}^{1}$

$Ni 28 \to [ Ar ] 3{d}^{8}4{s}^{2}; {Ni}^{2 +} \to [ Ar ] 3{d}^{8}$

$23 V \to [ Ar ] 3{d}^{3}4{s}^{2}; {V}^{+} \to [ Ar ] 3{d}^{3}4{s}^{1}$

Hence option B is the answer.

Match List-I with List-II List I (Species) List II (Electronic...