More practice questions

4. Q4. Write down the number of 3d electrons in each of these ions: Ti²⁺, V²⁺, Cr³⁺, Mn²⁺, Fe²⁺, Fe³⁺, Co²⁺, Ni²⁺ and Cu²⁺. (NCERT Exercise 4.36, first part.)

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   With no 4s electrons left, 3d electrons = Z − 18 − charge: Ti²⁺ 2, V²⁺ 3, Cr³⁺ 3, Mn²⁺ 5, Fe²⁺ 6, Fe³⁺ 5, Co²⁺ 7, Ni²⁺ 8, Cu²⁺ 9 (reasoned from NCERT page 95).

5. Q5. Write the electronic configuration of the element with atomic number 109. (Part of NCERT Exercise 4.34.)

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   Element 109 is meitnerium, in the 6d series: [Rn] 5f¹⁴ 6d⁷ 7s². NCERT Table 4.1 gives its outer configuration as 6d⁷ 7s² (page 91); it is a predicted configuration, as for all the heaviest elements (Wikipedia, “Electron configurations of the elements”).

6. Q6. For free ions or the usual high-spin aqueous ions, which has more unpaired electrons, Fe²⁺ or Fe³⁺?

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   Fe²⁺ is 3d⁶: five electrons enter the five d orbitals singly and the sixth pairs, leaving 4 unpaired. Fe³⁺ is 3d⁵, one electron in each d orbital: 5 unpaired. So high-spin Fe³⁺ has more; low-spin complexes require a separate ligand-field analysis (reasoned; Hund’s rule, NCERT Class 11 Chemistry, Unit 2).

7. Q7. Assertion: chromium is 3d⁵ 4s¹. Reason: the 3d and 4s orbitals are close in energy, and a half-filled 3d set is relatively more stable. Are both true, and does the reason explain the assertion?

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   Both are true, and the reason explains the assertion (NCERT page 90).

8. Q8. Molybdenum is 4d⁵ 5s¹, like chromium above it, but tungsten, below molybdenum, is 5d⁴ 6s². What does this show?

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   That the extra stability of a half-filled d set is a tendency, not a rule. It wins only when it outweighs the energy gap between the (n − 1)d and ns orbitals, and that balance differs from atom to atom (configurations from NCERT Table 4.1, page 91; reasoned).