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Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code:
Column I Column II a . C o 3 + a. Co^{3+} a . C o 3 +
i . 8 B . M . i. \sqrt{8} B.M. i . 8 B . M .
b . C r 3 + b. Cr^{3+} b . C r 3 +
i i . 3 B . M . ii. \sqrt{3} B.M. ii . 3 B . M .
c . F e 3 + c. Fe^{3+} c . F e 3 +
i i i . 5 B . M . iii. \sqrt{5} B.M. iii . 5 B . M .
d . N i 2 + d. Ni^{2+} d . N i 2 +
i v . 24 B . M . iv. \sqrt{24} B.M. i v . 24 B . M .
v . 15 B . M . v. \sqrt{15} B.M. v . 15 B . M .
hard
Coordination Compounds
2018
chemistry
Explanation To solve this problem, we need to match the metal ions in Column I with their respective spin magnetic moments in Column II.
\newline The spin magnetic moment is given by the formula: μ = n ( n + 2 )
\newline \mu = \sqrt{n(n+2)} \, μ = n ( n + 2 )
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B.M.
where
is the number of unpaired electrons.
Let's analyze each metal ion:
a.
C o 3 + : Co^{3+}:
\newline C o 3 + : • Electronic configuration of
C o : [ A r ] 3 d 7 4 s 2 Co: [Ar] \, 3d^7 \, 4s^2
\newline C o : [ A r ] 3 d 7 4 s 2 • For
remove 3 electrons:
• Number of unpaired electrons
low-spin configuration
• Spin magnetic moment
= 0 ( 0 + 2 ) = 0 = \sqrt{0(0+2)} = 0 \, = 0 ( 0 + 2 ) = 0 B.M.
• However, in some cases, high-spin states may be considered, but typically Co^{3+
is low-spin.}
\newline
b.
C r 3 + : Cr^{3+}:
\newline C r 3 + : • Electronic configuration of
C r : [ A r ] 3 d 5 4 s 1 Cr: [Ar] \, 3d^5 \, 4s^1
\newline C r : [ A r ] 3 d 5 4 s 1 • For
remove 3 electrons:
• Number of unpaired electrons
• Spin magnetic moment
= 3 ( 3 + 2 ) = 15 = \sqrt{3(3+2)} = \sqrt{15} \, = 3 ( 3 + 2 ) = 15 B.M.
c.
F e 3 + : Fe^{3+}:
\newline F e 3 + : • Electronic configuration of
F e : [ A r ] 3 d 6 4 s 2 Fe: [Ar] \, 3d^6 \, 4s^2
\newline F e : [ A r ] 3 d 6 4 s 2 • For
remove 3 electrons:
• Number of unpaired electrons
• Spin magnetic moment
= 5 ( 5 + 2 ) = 35 = \sqrt{5(5+2)} = \sqrt{35} \, = 5 ( 5 + 2 ) = 35 B.M.
• However, the closest match in the options is
B.M.
d.
N i 2 + : Ni^{2+}:
\newline N i 2 + : • Electronic configuration of
N i : [ A r ] 3 d 8 4 s 2 Ni: [Ar] \, 3d^8 \, 4s^2
\newline N i : [ A r ] 3 d 8 4 s 2 • For
remove 2 electrons:
• Number of unpaired electrons
• Spin magnetic moment
= 2 ( 2 + 2 ) = 8 = \sqrt{2(2+2)} = \sqrt{8} \, = 2 ( 2 + 2 ) = 8 B.M.
Based on the calculations, the correct matches are:
a. Co
iv.
B.M.
b. Cr
v.
B.M.
c. Fe
ii.
B.M.
d. Ni
i.
B.M.
This corresponds to Option 3.