Chemical Bonding & Molecular Structure
Which of the following is least ionic?
The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals
A. have the same energy
B. have the minimum overlap
C. have same symmetry about the molecular axis
D. have different symmetry about the molecular axis
Choose the most appropriate from the options given below:
Given below are two statements:
Statement - I: Since Fluorine is more electronegative than nitrogen, the net dipole moment of $\mathrm{NF}_3$ is greater than $\mathrm{NH}_3$.
Statement - II: In $\mathrm{NH}_3$, the orbital dipole due to lone pair and the dipole moment of $\mathrm{NH}$ bonds are in opposite direction, but in $\mathrm{NF}_3$ the orbital dipole due to lone pair and dipole moments of N-F bonds are in same direction.
In the light of the above statements, choose the most appropriate from the options given below:
The molecule / ion with square pyramidal shape is
Aluminium chloride in acidified aqueous solution forms an ion having geometry
Match List I with List II.
| List I Molecule |
List II Shape |
||
|---|---|---|---|
| (A) | $\mathrm{BrF_5}$ | (I) | T-Shape |
| (B) | $\mathrm{H_2O}$ | (II) | See saw |
| (C) | $\mathrm{ClF_3}$ | (III) | Bent |
| (D) | $\mathrm{SF_4}$ | (IV) | Square pyramidal |
Choose the correct answer from the options given below :
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : There is a considerable increase in covalent radius from $\mathrm{N}$ to $\mathrm{P}$. However from As to Bi only a small increase in covalent radius is observed.
Reason (R) : Covalent and ionic radii in a particular oxidation state increases down the group. In the light of the above statements, choose the most appropriate answer from the options given below:
The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is
Choose the polar molecule from the following:
Total number of electrons present in $\left(\pi^*\right)$ molecular orbitals of $\mathrm{O}_2, \mathrm{O}_2^{+}$ and $\mathrm{O}_2^{-}$ is ________.
Explanation:
$\begin{aligned} & \mathrm{O}_2(16 \mathrm{e}):\left(\sigma_{1 \mathrm{~s}}\right)^2\left(\sigma_{1 \mathrm{~s}}^*\right)^2\left(\sigma_{2 \mathrm{~s}}\right)^2\left(\sigma_{2 \mathrm{~s}}^*\right)^2 \\ & \left(\sigma_{2 \mathrm{p}}\right)^2\left[\left(\pi_{2 \mathrm{p}}\right)^2=\left(\pi_{2 \mathrm{p}}\right)^2\right],\left[\left(\pi_{2 \mathrm{p}}^*\right)^1=\left(\pi_{2 \mathrm{p}}^*\right)^1\right] \end{aligned}$
Number of $\mathrm{e}^{-}$ present in $\left(\pi^*\right)$ of $\mathrm{O}_2=2$
Number of $\mathrm{e}^{-}$ present in $\left(\pi^*\right)$ of $\mathrm{O}_2^{+}=1$
Number of $\mathrm{e}^{-}$ present in $\left(\pi^*\right)$ of $\mathrm{O}_2^{-}=3$So total $\mathrm{e}^{-}$ in $\left(\pi^*\right)=2+1+3=6$
The total number of species from the following in which one unpaired electron is present, is _______.
$\mathrm{N}_2, \mathrm{O}_2, \mathrm{C}_2^{-}, \mathrm{O}_2^{-}, \mathrm{O}_2^{2-}, \mathrm{H}_2^{+}, \mathrm{CN}^{-}, \mathrm{He}_2^{+}$
Explanation:
| Species | Unpaired e |
|---|---|
| $ \mathrm{N}_2 $ |
0 |
| $ \mathrm{O}_2 $ |
2 |
| $ \mathrm{C}_2^{-} $ |
1 |
| $ \mathrm{O}_2^{-} $ |
1 |
| $ \mathrm{O}_2^{2-} $ |
0 |
| $ \mathrm{H}_2^{+} $ |
1 |
| $ \mathrm{CN}^{-} $ |
0 |
| $ \mathrm{He}_2^{+} $ |
1 |
Number of molecules having bond order 2 from the following molecules is _________.
$\mathrm{C}_2, \mathrm{O}_2, \mathrm{Be}_2, \mathrm{Li}_2, \mathrm{Ne}_2, \mathrm{~N}_2, \mathrm{He}_2$
Explanation:
To determine the number of molecules with a bond order of 2 from the given molecules, we need to first calculate the bond order for each molecule. The bond order can be determined using Molecular Orbital Theory (MOT). The bond order is given by the formula:
$\text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of anti-bonding electrons}}{2}$
Let's evaluate each molecule individually:
1. $\mathrm{C}_2$:
For $\mathrm{C}_2$, the total number of electrons is 12. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2$. There are 8 bonding electrons and 4 anti-bonding electrons:
$\text{Bond Order} = \frac{8 - 4}{2} = 2$
2. $\mathrm{O}_2$:
For $\mathrm{O}_2$, the total number of electrons is 16. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \sigma_{2p_z} \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2 \left( \pi_{2p_x}^* \right)^1 \left( \pi_{2p_y}^* \right)^1$. There are 10 bonding electrons and 6 anti-bonding electrons:
$\text{Bond Order} = \frac{10 - 6}{2} = 2$
3. $\mathrm{Be}_2$:
For $\mathrm{Be}_2$, the total number of electrons is 8. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2$. There are 4 bonding electrons and 4 anti-bonding electrons:
$\text{Bond Order} = \frac{4 - 4}{2} = 0$
4. $\mathrm{Li}_2$:
For $\mathrm{Li}_2$, the total number of electrons is 6. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2$. There are 4 bonding electrons and 2 anti-bonding electrons:
$\text{Bond Order} = \frac{4 - 2}{2} = 1$
5. $\mathrm{Ne}_2$:
For $\mathrm{Ne}_2$, the total number of electrons is 20. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \sigma_{2p_z} \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2 \left( \pi_{2p_x}^* \right)^2 \left( \pi_{2p_y}^* \right)^2 \left( \sigma_{2p_z}^* \right)^2$. There are 10 bonding electrons and 10 anti-bonding electrons:
$\text{Bond Order} = \frac{10 - 10}{2} = 0$
6. $\mathrm{N}_2$:
For $\mathrm{N}_2$, the total number of electrons is 14. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \sigma_{2p_z} \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2$. There are 10 bonding electrons and 4 anti-bonding electrons:
$\text{Bond Order} = \frac{10 - 4}{2} = 3$
7. $\mathrm{He}_2$:
For $\mathrm{He}_2$, the total number of electrons is 4. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2$. There are 2 bonding electrons and 2 anti-bonding electrons:
$\text{Bond Order} = \frac{2 - 2}{2} = 0$
Thus, the molecules with a bond order of 2 from the given list are $\mathrm{C}_2$ and $\mathrm{O}_2$. Therefore, the number of molecules having bond order 2 is 2.
Number of molecules from the following which are exceptions to octet rule is _________.
$\mathrm{CO}_2, \mathrm{NO}_2, \mathrm{H}_2 \mathrm{SO}_4, \mathrm{BF}_3, \mathrm{CH}_4, \mathrm{SiF}_4, \mathrm{ClO}_2, \mathrm{PCl}_5, \mathrm{BeF}_2, \mathrm{C}_2 \mathrm{H}_6, \mathrm{CHCl}_3, \mathrm{CBr}_4$
Explanation:
$\mathrm{NO}_2, \mathrm{H}_2 \mathrm{SO}_4, \mathrm{BF}_3, \mathrm{ClO}_2, \mathrm{PCl}_5, \mathrm{BeF}_2$
These are exception of octet rule
Consider the following reactions
The number of protons that do not involve in hydrogen bonding in the product B is _________.
Explanation:
$\mathrm{NiS + HN{O_3} + HCl\buildrel {} \over \longrightarrow \mathop {NiC{l_2}}\limits_{(A)} + S + NO + {H_2}O}$
$\mathrm{\mathop {NiC{l_2}}\limits_{(A)} + N{H_4}OH +}$ Dimethylgyoxime $\mathrm{\buildrel {} \over \longrightarrow \mathop {Ni{{(dmg)}_2}}\limits_{(B)} + N{H_4}Cl + {H_2}O}$
Total 12 proton do not involve in H-bond
Total number of species from the following with central atom utilising $\mathrm{sp}^2$ hybrid orbitals for bonding is ________.
$\mathrm{NH}_3, \mathrm{SO}_2, \mathrm{SiO}_2, \mathrm{BeCl}_2, \mathrm{C}_2 \mathrm{H}_2, \mathrm{C}_2 \mathrm{H}_4, \mathrm{BCl}_3, \mathrm{HCHO}, \mathrm{C}_6 \mathrm{H}_6, \mathrm{BF}_3, \mathrm{C}_2 \mathrm{H}_4 \mathrm{Cl}_2$
Explanation:
$\mathrm{SO}_2, \mathrm{C}_2 \mathrm{H}_4, \mathrm{BCl}_3, \mathrm{HCHO}, \mathrm{C}_6 \mathrm{H}_6, \mathrm{BF}_3$ are $s p^2$ hybridised central atom
Number of molecules from the following which can exhibit hydrogen bonding is _________. (nearest integer)
Explanation:
compounds
$\mathrm{CH_3OH}$
$\mathrm{H_2O}$
$\mathrm{HF}$
$\mathrm{NH}_3$
Can show $\mathrm{H}$-Bonding.
Number of compounds from the following with zero dipole moment is _________.
$\mathrm{HF}, \mathrm{H}_2, \mathrm{H}_2 \mathrm{~S}, \mathrm{CO}_2, \mathrm{NH}_3, \mathrm{BF}_3, \mathrm{CH}_4, \mathrm{CHCl}_3, \mathrm{SiF}_4, \mathrm{H}_2 \mathrm{O}, \mathrm{BeF}_2$
Explanation:
A molecule has zero dipole moment when its bond dipoles cancel by symmetry (or when bonds are nonpolar).
| Species | Shape (VSEPR) | Dipole moment? | Reason |
|---|---|---|---|
| HF | linear | nonzero | polar bond, no cancellation |
| $\mathrm{H}_2$ | linear | zero | homonuclear, nonpolar |
| $\mathrm{H}_2\mathrm{S}$ | bent | nonzero | bent shape → net dipole |
| $\mathrm{CO}_2$ | linear | zero | symmetric linear cancellation |
| $\mathrm{NH}_3$ | trigonal pyramidal | nonzero | lone pair breaks symmetry |
| $\mathrm{BF}_3$ | trigonal planar | zero | symmetric planar cancellation |
| $\mathrm{CH}_4$ | tetrahedral | zero | symmetric tetrahedral cancellation |
| $\mathrm{CHCl}_3$ | tetrahedral | nonzero | not symmetric ($3\ \mathrm{Cl}$ and $1\ \mathrm{H}$) |
| $\mathrm{SiF}_4$ | tetrahedral | zero | symmetric tetrahedral cancellation |
| $\mathrm{H}_2\mathrm{O}$ | bent | nonzero | bent shape → net dipole |
| $\mathrm{BeF}_2$ | linear | zero | symmetric linear cancellation |
Compounds with zero dipole moment:
$\mathrm{H}_2,\ \mathrm{CO}_2,\ \mathrm{BF}_3,\ \mathrm{CH}_4,\ \mathrm{SiF}_4,\ \mathrm{BeF}_2$
Number = $6$.
In the lewis dot structure for $\mathrm{NO}_2^{-}$, total number of valence electrons around nitrogen is _________.
Explanation:
$\text { Lewis dot structure of } \mathrm{NO}_2^{-} \text {is: }$
$\text { Total number of valence } \mathrm{e}^{\ominus} \text { around } \mathrm{N}=8$
Number of compounds / species from the following with non-zero dipole moment is _________.
$\mathrm{BeCl}_2, \mathrm{BCl}_3, \mathrm{NF}_3, \mathrm{XeF}_4, \mathrm{CCl}_4, \mathrm{H}_2 \mathrm{O}, \mathrm{H}_2 \mathrm{~S}, \mathrm{HBr}, \mathrm{CO}_2, \mathrm{H}_2, \mathrm{HCl}$
Explanation:
$\mathrm{NF}_3, \mathrm{H}_2 \mathrm{O}, \mathrm{H}_2 \mathrm{S}, \mathrm{HBr}, \mathrm{HCl}$ have non zero dipole moments.
Number of molecules/species from the following having one unpaired electron is ________.
$\mathrm{O}_2, \mathrm{O}_2^{-1}, \mathrm{NO}, \mathrm{CN}^{-1}, \mathrm{O}_2^{2-}$
Explanation:
$\mathrm{O}_2^{-} \text {and } \mathrm{NO} \text { have } 1 \text { unpaired electron. }$
$\mathrm{PF}_5, \mathrm{BrF}_5, \mathrm{PCl}_5,\left[\mathrm{Pt} \mathrm{Cl}_4\right]^{2-}, \mathrm{BF}_3, \mathrm{Fe}(\mathrm{CO})_5$
Explanation:
Explanation:
When an atom has the lowest oxidation number of -2 in a compound, it means the atom has gained two electrons beyond its neutral state to achieve this oxidation state. This is because gaining electrons makes the oxidation number more negative. In a neutral atom, the electrons in the outermost shell are known as valence electrons, which play a crucial role in chemical reactions and bonding.
In the given compound $\mathrm{A}_2\mathrm{B}$, atom B has an oxidation state of -2, indicating it has gained two electrons. For an atom to gain two electrons to complete its octet implies that its neutral state had six valence electrons. Therefore, before gaining electrons, the atom B would have had six electrons in its valence shell to begin with. As it gains two more electrons, it reaches an oxidation number of -2, implying it now has a complete octet or eight electrons in its valence shell, but the original count of valence electrons in its neutral state was six.
A diatomic molecule has a dipole moment of $1.2 \mathrm{~D}$. If the bond distance is $1 \mathrm{~A}^{\circ}$, then fractional charge on each atom is _________ $\times 10^{-1}$ esu.
(Given $1 \mathrm{~D}=10^{-18}$ esucm)
Explanation:
To find the fractional charge on each atom in a diatomic molecule, we can use the relationship between dipole moment (μ), charge (q), and distance (d):
$ \mu = q \times d $
Where:
- μ is the dipole moment.
- q is the magnitude of the charge on each atom.
- d is the distance between the charges.
First, convert units to be consistent with esu:
- Bond distance in cm: $ 1 \mathrm{~A}^{\circ} = 1 \times 10^{-8} \mathrm{~cm} $
- Dipole moment in esu·cm: $ 1.2 \mathrm{~D} = 1.2 \times 10^{-18} \mathrm{~esucm} $
We are given that the bond distance ($ d $) is $ 1 \mathrm{~A}^{\circ} $ and that the dipole moment $( \mu $) is $ 1.2 \mathrm{~D} $. Using these values, we can calculate the charge ($ q $) as follows:
$ q = \frac{\mu}{d} $
Now plug in the given values:
$ q = \frac{1.2 \times 10^{-18} \mathrm{~esucm}}{1 \times 10^{-8} \mathrm{~cm}} $
$ q = 1.2 \times 10^{-10} \mathrm{~esu} $
$ q = 1.2 \times 10^{-9} \times 10^{-1} \mathrm{~esu} $
$ q = 0.000000012 \times 10^{-1} \mathrm{~esu} $
The number of species from the following in which the central atom uses $\mathrm{sp}^3$ hybrid orbitals in its bonding is __________.
$\mathrm{NH}_3, \mathrm{SO}_2, \mathrm{SiO}_2, \mathrm{BeCl}_2, \mathrm{CO}_2, \mathrm{H}_2 \mathrm{O}, \mathrm{CH}_4, \mathrm{BF}_3$
Explanation:
To determine the number of species in which the central atom uses $\mathrm{sp}^3$ hybrid orbitals in its bonding, we need to analyze the hybridization of each central atom in the given species.
1. $\mathrm{NH}_3$ (Ammonia):
The central atom is nitrogen. Ammonia has 3 sigma bonds and 1 lone pair. Thus, the steric number is 4, which corresponds to $\mathrm{sp}^3$ hybridization.
2. $\mathrm{SO}_2$ (Sulfur dioxide):
The central atom is sulfur. Sulfur dioxide has 2 sigma bonds and 1 lone pair. Thus, the steric number is 3, which corresponds to $\mathrm{sp}^2$ hybridization.
3. $\mathrm{SiO}_2$ (Silicon dioxide):
The central atom is silicon. Silicon dioxide has a linear structure with double bonds, leading to $\mathrm{sp}$ hybridization.
4. $\mathrm{BeCl}_2$ (Beryllium chloride):
The central atom is beryllium. Beryllium chloride has 2 sigma bonds and no lone pair. Thus, the steric number is 2, which corresponds to $\mathrm{sp}$ hybridization.
5. $\mathrm{CO}_2$ (Carbon dioxide):
The central atom is carbon. Carbon dioxide has a linear structure with double bonds, leading to $\mathrm{sp}$ hybridization.
6. $\mathrm{H}_2\mathrm{O}$ (Water):
The central atom is oxygen. Water has 2 sigma bonds and 2 lone pairs. Thus, the steric number is 4, which corresponds to $\mathrm{sp}^3$ hybridization.
7. $\mathrm{CH}_4$ (Methane):
The central atom is carbon. Methane has 4 sigma bonds and no lone pair. Thus, the steric number is 4, which corresponds to $\mathrm{sp}^3$ hybridization.
8. $\mathrm{BF}_3$ (Boron trifluoride):
The central atom is boron. Boron trifluoride has 3 sigma bonds and no lone pair. Thus, the steric number is 3, which corresponds to $\mathrm{sp}^2$ hybridization.
From the above analysis, the species in which the central atom uses $\mathrm{sp}^3$ hybrid orbitals are:
- $\mathrm{NH}_3$
- $\mathrm{H}_2\mathrm{O}$
- $\mathrm{CH}_4$
Therefore, the number of species where the central atom uses $\mathrm{sp}^3$ hybrid orbitals is 3.
The total number of molecular orbitals formed from $2 \mathrm{s}$ and $2 \mathrm{p}$ atomic orbitals of a diatomic molecule is __________.
Explanation:
Two molecular orbitals $\sigma 2 \mathrm{s}$ and $\sigma * 2 \mathrm{s}$.
Six molecular orbitals $\sigma 2 \mathrm{p}_z$ and $\sigma * 2 \mathrm{p}_{\mathrm{z}}$.
$\pi 2 \mathrm{p}_{\mathrm{x}}, \pi 2 \mathrm{p}_{\mathrm{y}}$ and $\pi * 2 \mathrm{p}_{\mathrm{x}}, \pi^* 2 \mathrm{p}_{\mathrm{y}}$
The total number of molecules with zero dipole moment among $\mathrm{CH}_4, \mathrm{BF}_3, \mathrm{H}_2 \mathrm{O}, \mathrm{HF}, \mathrm{NH}_3, \mathrm{CO}_2$ and $\mathrm{SO}_2$ is ________.
Explanation:
To determine which molecules have a zero dipole moment from the list provided, we need to consider their molecular geometry and symmetry. Molecules that are symmetrical often have zero dipole moment because the bond dipoles cancel each other out.
COâ‚‚ (Carbon Dioxide): This molecule is linear in shape. The two opposing oxygen atoms pull equally on the carbon, canceling out the dipoles, resulting in a net dipole moment of zero.
CHâ‚„ (Methane): Methane is tetrahedral and symmetrical, with all four hydrogen atoms equally spaced around the carbon atom. This symmetry leads to the cancellation of dipoles, giving a net dipole moment of zero.
BF₃ (Boron Trifluoride): This molecule is trigonal planar and symmetrical. The three fluorine atoms are arranged evenly around the central boron atom, causing the dipoles to cancel each other out, resulting in a zero dipole moment.
Therefore, the molecules with zero dipole moment among the list provided are CO₂, CH₄, and BF₃.
The total number of anti bonding molecular orbitals, formed from $2 s$ and $2 p$ atomic orbitals in a diatomic molecule is _______.
Explanation:
Antibonding molecular orbital from $2 \mathrm{~s}=1$
Antibonding molecular orbital from $2 p=3$
Total $=4$
The number of species from the following which are paramagnetic and with bond order equal to one is _________.
$\mathrm{H}_2, \mathrm{He}_2^{+}, \mathrm{O}_2^{+}, \mathrm{N}_2^{2-}, \mathrm{O}_2^{2-}, \mathrm{F}_2, \mathrm{Ne}_2^{+}, \mathrm{B}_2$
Explanation:
| Sol. | Magnetic behaviour | Bond order |
|---|---|---|
| $\mathrm{H}_2$ | Diamagnetic | $1$ |
| $\mathrm{He_2^+}$ | Paramagnetic | $0.5$ |
| $\mathrm{O_2^+}$ | Paramagnetic | $2.5$ |
| $\mathrm{N_2^{2-}}$ | Paramagnetic | $2$ |
| $\mathrm{O_2^{2-}}$ | Diamagnetic | $1$ |
| $\mathrm{F_2}$ | Diamagnetic | $1$ |
| $\mathrm{Ne_2^+}$ | Paramagnetic | $0.5$ |
| $\mathrm{B_2}$ | Paramagnetic | $1$ |
Number of compounds with one lone pair of electrons on central atom amongst following is _________.
$\mathrm{O}_3, \mathrm{H}_2 \mathrm{O}, \mathrm{SF}_4, \mathrm{ClF}_3, \mathrm{NH}_3, \mathrm{BrF}_5, \mathrm{XeF}_4$
Explanation:
The number of non-polar molecules from the following is _________. $\mathrm{HF}, \mathrm{H}_2 \mathrm{O}, \mathrm{SO}_2, \mathrm{H}_2, \mathrm{CO}_2, \mathrm{CH}_4, \mathrm{NH}_3, \mathrm{HCl}, \mathrm{CHCl}_3, \mathrm{BF}_3$
Explanation:
To determine whether a molecule is polar or non-polar, we must consider the difference in electronegativity between the atoms and the symmetry of the molecule. Polar molecules occur when there is an electronegativity difference between the bonded atoms. Non-polar molecules either do not have any polar bonds or the polarities cancel each other out because of a symmetrical arrangement. Let's consider each molecule individually:
- HF (Hydrogen fluoride): This molecule is polar due to the high electronegativity difference between hydrogen (H) and fluorine (F).
- H2O (Water): It is polar because of its bent shape and the difference in electronegativity between oxygen and hydrogen.
- SO2 (Sulfur dioxide): The molecule is non-linear and has polar bonds, making it a polar molecule.
- H2 (Hydrogen): This molecule is non-polar because it consists of two identical atoms which share electrons evenly.
- CO2 (Carbon dioxide): Even though C=O bonds are polar, the molecule is linear and the polarities cancel out, making CO2 non-polar.
- CH4 (Methane): It is non-polar because the C-H bonds are evenly distributed in a tetrahedral shape, canceling out any dipole moments.
- NH3 (Ammonia): This molecule is polar; it has a trigonal pyramidal shape with a lone pair on nitrogen, creating a dipole moment.
- HCl (Hydrogen chloride): The molecule is polar due to the electronegativity difference between hydrogen and chlorine.
- CHCl3 (Chloroform): It has polar C-Cl bonds and is not symmetrical, resulting in a polar molecule.
- BF3 (Boron trifluoride): This molecule is non-polar despite having polar bonds, because the shape of the molecule is trigonal planar and the dipoles cancel out.
Given this information, the non-polar molecules from the list are: $\mathrm{H}_2, \mathrm{CO}_2, \mathrm{CH}_4, \mathrm{BF}_3$.
Therefore, there are 4 non-polar molecules in the list given.
Sum of bond order of CO and NO$^+$ is ________.
Explanation:
$\begin{array}{lcl} \mathrm{CO} \Rightarrow & \overline{\mathrm{C}} \equiv \stackrel{+}{\mathrm{O}} & : \mathrm{BO}=3 \\ \mathrm{NO}^{+} \Rightarrow & \mathrm{N} \equiv \mathrm{O}^{+} & : \mathrm{BO}=3 \end{array}$
(A) NF3 molecule has a trigonal planar structure.
(B) Bond length of $\mathrm{N}_{2}$ is shorter than $\mathrm{O}_{2}$.
(C) Isoelectronic molecules or ions have identical bond order.
(D) Dipole moment of $\mathrm{H}_{2}\mathrm{S}$ is higher than that of water molecule.
Choose the correct answer from the options given below:
Given below are two statements :
Statement I : $\mathrm{SO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$ both possess V-shaped structure.
Statement II : The bond angle of $\mathrm{SO}_{2}$ is less than that of $\mathrm{H}_{2} \mathrm{O}$.
In the light of the above statements, choose the most appropriate answer from the options given below:
In which of the following processes, the bond order increases and paramagnetic character changes to diamagnetic one ?
The bond order and magnetic property of acetylide ion are same as that of
Match List - I with List - II:
| List - I Species | List - II Geometry/Shape | ||
|---|---|---|---|
| A. | $\mathrm{H_3O^+}$ | I. | Tetrahedral |
| B. | Acetylide anion | II. | Linear |
| C. | $\mathrm{NH_4^+}$ | III. | Pyramidal |
| D. | $\mathrm{ClO_2^-}$ | IV. | Bent |
Choose the correct answer from the options given below:
The compound which does not exist is
Match List I with List II
| LIST I Oxide |
LIST II Type of bond |
||
|---|---|---|---|
| A. | $\mathrm{N_2O_4}$ | I. | 1 N = O bond |
| B. | $\mathrm{NO_2}$ | II. | 1 N $-$ O $-$ N bond |
| C. | $\mathrm{N_2O_5}$ | III. | 1 N $-$ N bond |
| D. | $\mathrm{N_2O}$ | IV. | 1 N=N / N $\equiv$ N bond |
Choose the correct answer from the options given below:
$\mathrm{O}-\mathrm{O}$ bond length in $\mathrm{H}_{2} \mathrm{O}_{2}$ is $\underline{\mathrm{X}}$ than the $\mathrm{O}-\mathrm{O}$ bond length in $\mathrm{F}_{2} \mathrm{O}_{2}$. The $\mathrm{O}-\mathrm{H}$ bond length in $\mathrm{H}_{2} \mathrm{O}_{2}$ is $\underline{Y}$ than that of the $\mathrm{O}-\mathrm{F}$ bond in $\mathrm{F}_{2} \mathrm{O}_{2}$.
Choose the correct option for $\underline{X}$ and $\underline{Y}$ from those given below :
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | $\mathrm{XeF_4}$ | I. | See-saw |
| B. | $\mathrm{SF_4}$ | II. | Square-planar |
| C. | $\mathrm{NH_{4}^{+}}$ | III. | Bent T-shaped |
| D. | $\mathrm{BrF_3}$ | IV. | Tetrahedral |
Choose the correct answer from the options given below :
For $\mathrm{OF}_{2}$ molecule consider the following :
A. Number of lone pairs on oxygen is 2 .
B. FOF angle is less than $104.5^{\circ}$.
C. Oxidation state of $\mathrm{O}$ is $-2$.
D. Molecule is bent '$\mathrm{V}$' shaped.
E. Molecular geometry is linear.
correct options are:
Match List I with List II
| List I (molecules/ions) |
List II (No. of lone pairs of e$^-$ on central atom) |
||
|---|---|---|---|
| A. | $\mathrm{IF_7}$ | I. | Three |
| B. | $\mathrm{ICl}$$_4^ - $ | II. | One |
| C. | $\mathrm{XeF_6}$ | III. | Two |
| D. | $\mathrm{XeF_2}$ | IV. | Zero |
Choose the correct answer from the options given below :
According to MO theory the bond orders for $\mathrm{O}$$_2^{2 - }$, $\mathrm{CO}$ and $\mathrm{NO^+}$ respectively, are
The magnetic behaviour of $\mathrm{Li_2O,Na_2O_2}$ and $\mathrm{KO_2}$, respectively, are :
The bond dissociation energy is highest for
Statement I : Dipole moment is a vector quantity and by convention it is depicted by a small arrow with tail on the negative centre and head pointing towards the positive centre.
Statement II : The crossed arrow of the dipole moment symbolizes the direction of the shift of charges in the molecules.
In the light of the above statements, choose the most appropriate answer from the options given below :
What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species : $\mathrm{{N_2};N_2^ + ;{O_2};O_2^ + }$ ?