MCQ practice · SSC Chemistry Board questions · Chapter 3: Structure of Matter
For ⁵⁶₂₆Fe²⁺ −
- i. the neutron number is 30
- ii. it acts as a reducing agent
- iii. the electronic configuration is [Ar] 3d⁶4s²
Think it through first, then pick an option.
- a i and ii
- b ii and iii
- c i and iii
- d i, ii and iii
Answer and explanation
Correct answer: a i and ii
Right. i: neutrons = 56 − 26 = 30 (an ion keeps its neutrons). ii: Fe²⁺ can give up one more electron to reach the stable 3d⁵ (Fe³⁺), so it is a reducing agent. iii is false: ...3d⁶ 4s² belongs to the Fe atom; Fe²⁺ has lost 4s².
Why the other options are wrong
- b ii and iii: iii is false: Fe²⁺ has no 4s², it is ...3d⁶. And the true i is left out.
- c i and iii: iii is false: ...3d⁶ 4s² is the Fe atom, not Fe²⁺. And the true ii is left out.
- d i, ii and iii: iii is false: the two outermost 4s² electrons leave first, so Fe²⁺ is ...3d⁶.
More board questions on this topic
- The A²⁺ ion of the stimulus −
- How many electrons are in the M shell of ₂₅Mn²⁺?
- The electronic configuration 1s²2s²2p⁶3s²3p⁶ is seen in −
- Which element follows the (n − 1)d¹⁰ ns¹ configuration?
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