JEE Main 1-Sep-2021 Shift 2 Solved Paper

Spin only mangnetic moment is given as
μS=√n(n+2)‌BM
where, n = number of unpaired electrons
Electronic configuration of mathrm‌B2+is as follows
B2+=σ1s2,σ*1s2,σ2s2,σ*2s2,π2py1 =π2pz0
∴ B2+ has one unpaired electron,
μ=√1(1+2)=√3=1.73‌BM
=173×10−2‌BM
∴x×10−2=173×10−2
∴ x = 173
Hence, answer is 173.