The Cl 2 molecules in crystalline chlorine are arranged in herringbone arrays within sheets, held together by covalent bonds. The shortest Cl to Cl distance within these sheets is 1.98 mathringA . Adjacent sheets are held together by van der Waals' forces and the closest approach between Cl atoms on neighboring sheets is 3.69 mathringA . In solid HCl observed at lower temperatures, the H - Cl bond length is 1.25 mathringA. Therefore shortest distance between two Cl atoms in solid HCl is (given van der Waals' radius of H =1.1 mathringA ):

Question

The Cl 2 molecules in crystalline chlorine are arranged in herringbone arrays within sheets, held together by covalent bonds. The shortest Cl to Cl distance within these sheets is 1.98 mathringA . Adjacent sheets are held together by van der Waals' forces and the closest approach between Cl atoms on neighboring sheets is 3.69 mathringA . In solid HCl observed at lower temperatures, the H - Cl bond length is 1.25 mathringA. Therefore shortest distance between two Cl atoms in solid HCl is (given van der Waals' radius of H =1.1 mathringA ): (A) 2.5 mathringA (B) 2.95 mathringA (C) 4.2 mathringA (D) 6.04 mathringA

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/700ff3fd2e96e017f19745cfd5ba07ad-.png" /> d Cl - Cl =d H - Cl +r v . d ( H )+r v .d ( Cl ) =1.25+1.1+1.85=4.2 mathringA

$2.5 \overset{˚}{A}$

$2.95 \overset{˚}{A}$

$4.2 \overset{˚}{A}$

$6.04 \overset{˚}{A}$

$d_{Cl - Cl} = d_{H - Cl} + r_{v . d} (H) + r_{v . d} (Cl)$

$= 1.25 + 1.1 + 1.85 = 4.2 \overset{˚}{A}$

2.5A˚

2.95A˚

4.2A˚

6.04A˚

dCl−Cl​=dH−Cl​+rv.d​(H)+rv.d​(Cl) =1.25+1.1+1.85=4.2A˚

$2.5 \overset{˚}{A}$

$2.95 \overset{˚}{A}$

$4.2 \overset{˚}{A}$

$6.04 \overset{˚}{A}$

$d_{Cl - Cl} = d_{H - Cl} + r_{v . d} (H) + r_{v . d} (Cl)$

$= 1.25 + 1.1 + 1.85 = 4.2 \overset{˚}{A}$