MCQ
The ratio between radii of $H{e^ + }$ ion and $H$ atom is
- A$0.5$
- ✓$1$
- C$1.5$
- D$2$
Value of $Z$ for helium = $2$
Value of $n$ for both is = $1$
${r_H} = \frac{{0.52 \times {1^2}}}{1}$
${r_{H{e^ + }}} = \frac{{0.52 \times {1^2}}}{1}$$\frac{{{r_H}}}{{{r_{H{e^ + }}}}} = 1:1$ or ${r_{H{e^ + }}}:{r_H} = 1:1$
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$STATEMENT$-$2$: In water, orthoboric acid acts as a proton donor.
