Among the following series of transition metal ions, the one where all metal ions have 3d2 electronic configuration is [At. Nos. Ti = 22, V = 23, Cr = 24, Mn = 25]
- Ti3+, V2+, Cr3+, Mn4+
- Ti+, V4+, Cr6+, Mn7+
- Ti4+, V3+, Cr2+, Mn3+
- Ti2+, V3+, Cr4+, Mn5+
Answer
(4) Ti2+, V3+, Cr4+, Mn5+
Solution
When a 3d-series atom forms an ion, the 4s electrons are lost first.
Number of 3d electrons in Mn+ = Z - 18 - n
Ti2+: 22 - 18 - 2 = 2; V3+: 23 - 18 - 3 = 2
Cr4+: 24 - 18 - 4 = 2; Mn5+: 25 - 18 - 5 = 2
All ions of series (4) are 3d2 (in (1), Ti3+ is 3d1; in (2), Ti+ is 3d24s1 and V4+ is 3d1; in (3), Ti4+ is 3d0).
Answer: (4)