Question

the radius of 2nd orbit in B+4 ions is:

Answer 1

★$\color{darkblue}\underline{\underline{\sf To \: Find-}}$

• Radius of 2nd orbit of ${\sf B^{+4}}$ ion

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★$\color{darkblue}\underline{\underline{\sf Formula \: Used-}}$

$\color{violet}\bullet\underline{\boxed{\sf r=0.529×\dfrac{n^2}{Z}\:A°}}$

n = orbital

Z = atomic number

$\underline\color{red}{\bf NOTE}$

This formula is only applicable for Hydrogen and hydrogen like species i.e. species containing single electron.

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★$\color{darkblue}\underline{\underline{\sf Solution-}}$

Orbital (n) 2

Atomic number of boron (B) is 5

$\implies{\sf r=0.529×\dfrac{2^2}{5}}$

$\implies{\sf r=0.529×\dfrac{4}{5} }$

$\color{red}\implies{\sf r=0.42\:A° }$

★$\color{darkblue}\underline{\underline{\sf Answer-}}$

Radius of 2nd orbit of ${\sf B^{+4}}$ ion is $\color{red}{\sf 0.42\:A°}$