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Question Number 17209 by sushmitak last updated on 02/Jul/17

$$\mathrm{Spin}\mathrm{only}\mathrm{magnetic}\mathrm{moment}$$$$\mathrm{of}{}_{25}{\mathrm{Mn}}^{\mathrm{x}+}\mathrm{is}\sqrt{15}\mathrm{B}.\mathrm{M}.\mathrm{Then}$$$$\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{is}?$$$$\mathrm{i}\mathrm{did}\mathrm{following}$$$${1\mathrm{s}}^2{2\mathrm{s}}^2{2\mathrm{p}}^6{3\mathrm{s}}^2{3\mathrm{p}}^6{4\mathrm{s}}^2{3\mathrm{d}}^5$$$$\mathrm{so}\mathrm{to}\mathrm{get}3\mathrm{unpaired}\mathrm{electron}$$$$\mathrm{we}\mathrm{need}\mathrm{to}2\mathrm{electron}\mathrm{so}\mathrm{x}=2.$$$$\mathrm{book}\mathrm{says}\mathrm{x}=4.\mathrm{Why}?$$

Answered by Tinkutara last updated on 02/Jul/17

$${\mathrm{Book}}^{\prime }\mathrm{s}\mathrm{answer}\mathrm{is}\mathrm{correct}.$$$$\left[\mathrm{Mn}\right]={}_{18}\left[\mathrm{Ar}\right]4s^{2}3d^5$$$$\mathrm{and}\mathrm{magnetic}\mathrm{moment}=\sqrt{15}\mathrm{BM}$$$$\mathrm{So}\mathrm{unpaired}\mathrm{electrons}=3$$$$\mathrm{Losing}\mathrm{first}2\mathrm{electrons}\mathrm{from}4s\mathrm{orbital}$$$$\mathrm{and}\mathrm{next}2\mathrm{from}3d\mathrm{orbital}\mathrm{will}\mathrm{result}$$$$\mathrm{in}{}_{18}\left[\mathrm{Ar}\right]3d^3\mathrm{which}\mathrm{satisfies}\mathrm{that}$$$$\mathrm{number}\mathrm{of}\mathrm{unpaired}\mathrm{electrons}\mathrm{is}3.$$$$\mathrm{Hence}\mathrm{it}\mathrm{is}{\mathrm{Mn}}^{+4}.$$

Commented by sushmitak last updated on 02/Jul/17

$$\mathrm{Thank}\mathrm{You}.$$$$\mathrm{So}3\mathrm{d}\mathrm{fills}\mathrm{first}\mathrm{but}4\mathrm{s}\mathrm{is}\mathrm{first}\mathrm{to}$$$$\mathrm{lose}\mathrm{during}\mathrm{ionization}?$$

Commented by Tinkutara last updated on 02/Jul/17

$$\mathrm{Yes},\mathrm{exactly}.$$

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