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Which of the following has the regular tetrahedral structure ?
The [BFâ‚„]â» ion has a regular tetrahedral structure. Boron is surrounded by four fluorine atoms with no lone pairs on the boron atom, giving it a perfect tetrahedral shape. The other options do not have a regular tetrahedral geometry.
In $OF_2$ , number of bond pairs and lone pairs of electrons are respectively
Bond strength increases with
Bond strength is directly related to bond order. Bond order refers to the number of chemical bonds between a pair of atoms. Higher bond order means more electrons are shared between the atoms, leading to a stronger bond. Therefore, bond strength increases with bond order.
$ O_2 ^ {2+ } $ will have
In a molecule number of electrons in BMO are more as compared to ABMO, hence
In molecular orbital theory, if the number of electrons in bonding molecular orbitals (BMO) is greater than the number of electrons in anti-bonding molecular orbitals (ABMO), a bond will be formed. This is because the net stabilization energy is greater than the destabilization energy, leading to a stable bond.
The bond angle in the ammonium ion is equal to
The ammonium ion (NHâ‚„âº) has a tetrahedral geometry due to sp³ hybridization, which results in bond angles of approximately 109.28°.
The correct order of dipole moment is
The correct order of the O – O bond length in $O_2 , H_2O_2 and O_3$ is
The bond length of a molecule is inversely proportional to the bond order. The bond orders for $O_2$, $H_2O_2$, and $O_3$ are 2, 1, and 1.5 respectively. Therefore, the bond length order is: $H_2O_2 extgreater O_3 extgreater O_2$.
Choose the incorrect statement.
A sigma (σ) bond is generally stronger than a pi (π) bond. The sigma bond is formed by the end-to-end overlap of atomic orbitals, which allows for a stronger overlap compared to the side-to-side overlap of the pi bond. Therefore, the statement 'σ bond is weaker than π bond' is incorrect.
Which of the following is not paramagnetic ?
$S^{2-}$ (sulfide ion) is not paramagnetic because it has no unpaired electrons. Paramagnetism arises due to the presence of unpaired electrons, which are absent in $S^{2-}$. Hence, $S^{2-}$ is diamagnetic.
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