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Question
The metal complex ion that is paramagnetic is ______.
(Atomic number of Fe = 26, Cu = 29, Co = 27 and Ni = 28)
Options
[Fe(CN)4]2−
[Co(NH3)6]3+
[Ni(CN4)]2−
[Cu(NH3)4]2+
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Solution
The metal complex ion that is paramagnetic is [Cu(NH3)4]2+.
Explanation:
Paramagnetic substances have unpaired electrons, while diamagnetic substances have all electrons paired
- [Fe(CN)4]2−
Assume x is the oxidation number of Fe in given complex.
x + 4(−1) = −2
x − 4 = − 2
x = − 2 + 4
= + 2
Electronic configuration of Fe = [Ar]3d64s2
Fe+2 = [Ar]3d6
All electrons are paired; hence, it is diamagnetic in nature [Co(NH3)6]3+ - Assume x is the oxidation number of Co in given complex.
x + 6(0) = + 3
x = + 3
Electronic configuration of Co = [Ar] 3d74s2
Co+3 = [Ar]3d6 - All electrons are paired; hence, it is diamagnetic [Ni(CN4)]2−
Assume x is the oxidation number of Ni in the given complex.
x + 4(−1) =− 2
x − 4 = − 2
x = + 2
Electronic configuration of Ni = [Ar] 3d84s2
Ni2+ = [Ar]3d8 - All electrons are paired, hence, it is diamagnetic [Cu(NH3)4]2+.
Assume x is the oxidation number of Cu in the given complex.
x + 4(0) = + 2
x = + 2
Electronic configuration of Cu = [Ar] 3d10 4s1
Cu−2 = [Ar]3d9
It has one unpaired electron; hence, it is paramagnetic.
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