For each of the questions, four choices have been provided. Select the correct alternative. \(\mathrm{NH}^{+}{ }_{4}\) is isostructural with (a) \(\mathrm{PCl}_{3}\) (b) \(\mathrm{CH}_{4}\) (c) \(\mathrm{BF}_{3}\) (d) \(\mathrm{NO}_{3}^{-}\)

Short Answer

Expert verified
(a) \(\mathrm{PCl}_{3}\), (b) \(\mathrm{CH}_{4}\), (c) \(\mathrm{BF}_{3}\), (d) \(\mathrm{NO}_{3}^{-}\) Answer: (b) \(\mathrm{CH}_{4}\)

Step by step solution

01

Determine the shape of \(\mathrm{NH}^{+}{ }_{4}\)

The central nitrogen atom in \(\mathrm{NH}^{+}{ }_{4}\) has 5 valence electrons and shares 4 electrons with hydrogen atoms. It forms 4 single N-H covalent bonds. There are no lone pairs on the central nitrogen atom in this ion. As there are 4 electron pairs (bonding) around the central nitrogen atom that are arranged in a tetrahedron shape to minimize electron repulsion, the molecule has a tetrahedral geometry.
02

Determine the shape of \(\mathrm{PCl}_{3}\)

The central phosphorus atom in \(\mathrm{PCl}_{3}\) has 5 valence electrons and shares 3 across single P-Cl covalent bonds. The other two valence electrons remain as a lone pair on the central Phosphorus atom. Arranging the 3 bonding electron pairs and 1 lone pair as far apart as possible to minimize electron repulsion, we get a trigonal pyramidal shape for \(\mathrm{PCl}_{3}\).
03

Determine the shape of \(\mathrm{CH}_{4}\)

The central carbon atom in \(\mathrm{CH}_{4}\) has 4 valence electrons and shares all of them across single C-H covalent bonds. There are no lone pairs on the central carbon atom in this molecule. As there are 4 electron pairs (bonding) around the central carbon atom that are arranged in a tetrahedron shape to minimize electron repulsion, the molecule has a tetrahedral geometry.
04

Determine the shape of \(\mathrm{BF}_{3}\)

The central boron atom in \(\mathrm{BF}_{3}\) has 3 valence electrons and shares all of them across single B-F covalent bonds. There are no lone pairs on the central boron atom in this molecule. As there are 3 electron pairs (bonding) around the central boron atom that are arranged in a trigonal planar shape to minimize electron repulsion, the molecule has a trigonal planar geometry.
05

Determine the shape of \(\mathrm{NO}_{3}^{-}\)

The \(\mathrm{NO}_{3}^{-}\) ion is a resonance hybrid of three resonance structures with the same geometry. The central nitrogen atom is surrounded by 3 electron pairs (bonding) and one extra electron (which gives the ion its negative charge). As there are 3 electron pairs arranged in a trigonal planar shape to minimize electron repulsion, the molecule has a trigonal planar geometry.
06

Find the isostructural molecule

Based on the analyses in the previous steps, we know that \(\mathrm{NH}^{+}{ }_{4}\) and \(\mathrm{CH}_{4}\) both have tetrahedral geometries. Hence, \(\mathrm{NH}^{+}{ }_{4}\) is isostructural with \(\mathrm{CH}_{4}\). The correct alternative is (b) \(\mathrm{CH}_{4}\).

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