The number of molecules or ions from the following, which do not have odd number of electrons are _________.
(A) NO₂
(B) ICl₄ -
(C) BrF₃
(D) ClO₂
(E) NO₂ +
(F) NO

Solution

Chemistry ma’am · 3 steps

Add up the valence electrons, adjust for the charge, and check each total for odd or even.

1. Count valence electrons. Add the valence electrons of every atom. For an anion add one electron per negative charge, and for a cation take one away. Counting all electrons instead gives the same odd or even result, because inner shells are always full and even.

   Species	Valence electrons	Odd or even
   (A) NO₂	5 + 2(6) = 17	odd
   (B) ICl₄⁻	7 + 4(7) + 1 = 36	even
   (C) BrF₃	7 + 3(7) = 28	even
   (D) ClO₂	7 + 2(6) = 19	odd
   (E) NO₂⁺	5 + 2(6) - 1 = 16	even
   (F) NO	5 + 6 = 11	odd

2. Role of the charge. The charge changes the answer for both ions. Without it, ICl₄⁻ would have 35 and NO₂⁺ would have 17, and both totals are odd.
3. So. ICl₄⁻, BrF₃ and NO₂⁺ have an even number of electrons, so the answer is 3.

The common slip: 1. Ignoring the charges on ICl₄⁻ and NO₂⁺ makes both look odd, which gives 1 instead of 3.

The idea to keep. To spot an odd-electron species, sum the valence electrons and always apply the charge. Removing or adding one electron turns odd into even, so NO₂ is odd while NO₂⁺ is even.

Answer from the NTA final answer key. Solution by PrepWiser.