NEET Solved Paper 2018 Question 24
Question: 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: [NEET - 2018]
Column I Column II a. $ C{o^{\text{3+}}} $ i. $ \sqrt{8}\text{ BM} $ b. $ C{r^{\text{3+}}} $ ii. $ \sqrt{35}BM $ c. $ F{e^{\text{3+}}} $ iii. $ \sqrt{3}BM $ d. $ N{i^{\text{2+}}} $ iv. $ \sqrt{24}BM $ v. $ \sqrt{15}BM $
Options:
A) a-iv b-i c-ii d-iii
B) a-I b-ii c-iii d-iv
C) a-iv b-v c-ii d-i
D) a-iii b-v c-I d-ii
Show Answer
Answer:
Correct Answer: C
Solution:
- $ C{o^{\text{3+}}}\text{= }[\text{ Ar }]\text{ 3}{d^{6}} $ , Unpaired $ {e^{–}}\text{(n)=4} $ Spin magnetic moment $ \text{=}\sqrt{4(4+2)}=\sqrt{24}BM $ $ C{r^{\text{3+}}}\text{= }[\text{ Ar }]\text{ 3}{d^{3}}\text{,} $ Unpaired $ {e^{-}}\text{(n)=3} $ Spin magnetic moment $ \text{=}\sqrt{3(3+2)}=\sqrt{15}BM $ $ Fe^{3}+=[Ar]3d^{5} $ , Unpaired $ {e^{}}(n)=5 $ Spin magnetic moment $ =\sqrt{5(5+2)}=\sqrt{35}BM $ $ N{i^{\text{2+}}}\text{= }[\text{ Ar }]\text{ 3}{d^{8}}\text{,} $ Unpaired $ {e^{-}}(n)=2 $ Spin magnetic moment $ =\sqrt{2(2+2)}=\sqrt{8}BM $
