\(XeF_4\)
\(XeOF_2\)
\(XeO_2F_2\)
\(XeO_4\)
\(XeF_4 \) has a square planar molecular geometry. The central xenon \((Xe)\) atom is bonded to four fluorine \((F)\) atoms, and the molecule lies in a single plane. This arrangement results in a planar structure for \(XeF_4 \).
In contrast, the other compounds have different molecular geometries. \(XeOF_2\) has a bent shape, \(XeO_2F_2\) has a seesaw shape, and \(XeO_4\) has a tetrahedral shape.
So, the correct option is (A): \(XeF_4 \)
Identify the correct orders against the property mentioned:
A. H$_2$O $>$ NH$_3$ $>$ CHCl$_3$ - dipole moment
B. XeF$_4$ $>$ XeO$_3$ $>$ XeF$_2$ - number of lone pairs on central atom
C. O–H $>$ C–H $>$ N–O - bond length
D. N$_2$>O$_2$>H$_2$ - bond enthalpy
Choose the correct answer from the options given below:
Identify the suitable reagent for the following conversion: $Ph-C(=O)-OCH_3$ $\longrightarrow$ $Ph-CHO$
Identify the correct orders against the property mentioned:
A. H$_2$O $>$ NH$_3$ $>$ CHCl$_3$ - dipole moment
B. XeF$_4$ $>$ XeO$_3$ $>$ XeF$_2$ - number of lone pairs on central atom
C. O–H $>$ C–H $>$ N–O - bond length
D. N$_2$>O$_2$>H$_2$ - bond enthalpy
Choose the correct answer from the options given below:
| List - IMolecule | List - IIBond enthalpy (kJ mol-1) |
|---|---|
| (A) HCl | (I) 435.8 |
| (B) N2 | (II) 498 |
| (C) H2 | (III) 946.0 |
| (D) O2 | (IV) 431.0 |







Such a group of atoms is called a molecule. Obviously, there must be some force that holds these constituent atoms together in the molecules. The attractive force which holds various constituents (atoms, ions, etc.) together in different chemical species is called a chemical bond.
There are 4 types of chemical bonds which are formed by atoms or molecules to yield compounds.