The ionic radii of $S^{2 -} a n d T e^{2 -}$ are 1.84 $A^{0}$ and 2.2 $A^{0}$ respectively. The expected ionic radius of $S e^{2 -}$ is:

2.025

A^{0}

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1.82

A^{0}

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2.34

A^{0}

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None of these

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Solution

The correct option is B $2.025$ $A^{0}$
Ionic radius of $S e^{2 -}$ is expected to be in between the ionic radii of $S^{2 -}$ and $T e^{2 -} .$ Thus Ionic radius of $S e^{2 -} = \frac{1.84 + 2.21}{2} = 2.025 A^{0}$

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If Avogadro's number is A $_{0}$ , the number of sulphur atoms present in 200 mL of 1N H $_{2}$ SO $_{4}$ is

A_{0} A_{1} A_{2} A_{3} A_{4} A_{5}

A_{0} A_{1}, A_{0} A_{2}

A_{0} A_{4}

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A_{0} A_{1}, A_{0} A_{2}

A_{0} A_{4}

\to

\frac{1}{a_{0}} - \frac{x}{a_{0} a_{1}} + \frac{x^{2}}{a_{0} a_{1} a_{2}} - \dots + {(- 1)}^{n} \frac{x^{n}}{a_{0} a_{1} a_{2} \dots a_{n}}

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