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Question 96 of 119

Q.Write four points of differences between properties of nitrogen and other elements of group 15. Explain the structure of ClF3ClF_3. OR Conductivity of a solution is 6.23×10−5 Ω−1cm−16.23 \times 10^{-5}\ \Omega^{-1}cm^{-1} and its resistance is 13710 Ω. If the electrodes are 0.7 cm apart, calculate the cross-sectional area of the electrode. Why is molality of a solution independent of the temperature?

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2016Subjective· 7mImportance★★★★★
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Answering either OR alternative: (1) N vs Group 15 + ClF3ClF_3 shape, or (2) conductivity-cell area + why molality is T-independent.

Option 1 — Nitrogen vs other Group 15 elements (any four differences):

  1. Nitrogen exists as a diatomic gas (N2N_2, with a strong N≡NN\equiv N triple bond) at room temperature, while P, As, Sb, Bi are solids.
  2. Nitrogen shows a maximum covalency of 4 (limited by the absence of d-orbitals in its valence shell), whereas the heavier members can expand their covalency up to 5 or 6 using available d-orbitals.
  3. Because of this, nitrogen does not form pentahalides (e.g. no NCl5NCl_5), while P, As, Sb readily form pentahalides such as PCl5PCl_5.
  4. Nitrogen shows very weak catenation (N–N single bonds are weak due to lone-pair–lone-pair repulsion), whereas phosphorus shows strong catenation, forming chains and rings (e.g. P4P_4).
  5. Nitrogen, being small and highly electronegative, forms strong pπp\pi–pπp\pi multiple bonds, while heavier congeners prefer single bonds or dπd\pi–pπp\pi bonding.

Structure of ClF3ClF_3: the central Cl atom is sp3dsp^3d hybridised, with 3 Cl–F bond pairs and 2 lone pairs occupying the equatorial positions of a trigonal bipyramidal electron arrangement (lone pairs are placed equatorially to minimise repulsion). The resulting molecular shape is T-shaped (bent), with F–Cl–F bond angles slightly less than 90° due to lone pair–bond pair repulsions.

Option 2 (OR) — Conductivity cell area + molality:

Given κ=6.23×10−5 Ω−1cm−1\kappa = 6.23\times10^{-5}\ \Omega^{-1}cm^{-1}, R=13710 ΩR = 13710\ \Omega, l=0.7 cml = 0.7\ cm.

Cell constant G∗=l/A=κ×RG^* = l/A = \kappa \times R

κR=6.23×10−5×13710=0.8541\kappa R = 6.23\times10^{-5} \times 13710 = 0.8541 …

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