The atomic mass of an element is 133 and number of neutrons is 41.8% more than number of protons. Find the number of unpaired electrons.

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Solution

Atomic mass = no. of protons + no. of neutrons = n + p
Given that; $n = p + (\frac{41.8}{100}) p$
Therefore 133 = $n = p + p + (\frac{41.8}{100}) p$
133 = $\frac{100 p + 100 p + 41.8 p}{100}$
Therefore p = $\frac{133 \times 100}{241.8}$ = 55
Therefore electronic configuration = $[X e] 6 s^{1}$
No. of unpaired electrons = 1.

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The atomic mass of an element is 133 and number of neutrons is 41.8% more than number of protons. Find the number of unpaired electrons.

Atomic mass = no. of protons + no. of neutrons = n + pGiven that; n=p+(41.8100)pTherefore 133 = n=p+p+(41.8100)p133 = 100p+100p+41.8p100Therefore p = 133×100241.8 = 55Therefore electronic configuration = [Xe]6s1No. of unpaired electrons = 1.