MCQ
Electronic configuration of sodium is:
- A(2, 8, 1)
- B(2, 8, 2)
- C(2, 8, 3 )
- D(2, 8, 4)
Explanation:
Atomic no. of Na is 11 & thus the E.C is: 2, 8, 1
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${A_2}(g) + 4{B_2}(g) \rightleftharpoons 2A{B_4}(g)\,;\,\Delta H < 0$
the formation of $AB_4(g)$ will be favoured by