Chemical Bonding & Molecular Structure
Decreasing order of the hydrogen bonding in following forms of water is correctly represented by
A. Liquid water
B. Ice
C. Impure water
Choose the correct answer from the options given below :
Order of Covalent bond;
$\mathrm{A.~KF > KI ; LiF > KF}$
$\mathrm{B.~KF < KI ; LiF > KF}$
$\mathrm{C.~SnCl_4 > SnCl_2 ; CuCl > NaCl}$
$\mathrm{D.~LiF > KF ; CuCl < NaCl}$
$\mathrm{E.~KF < KI ; CuCl > NaCl}$
Choose the correct answer from the options given below :
The maximum number of lone pairs of electrons on the central atom from the following species is ____________.
$\mathrm{ClO}_{3}{ }^{-}, \mathrm{XeF}_{4}, \mathrm{SF}_{4}$ and $\mathrm{I}_{3}{ }^{-}$
Explanation:
The number of molecules from the following which contain only two lone pair of electrons is ________
$\mathrm{H}_{2} \mathrm{O}, \mathrm{N}_{2}, \mathrm{CO}, \mathrm{XeF}_{4}, \mathrm{NH}_{3}, \mathrm{NO}, \mathrm{CO}_{2}, \mathrm{~F}_{2}$
Explanation:
The number of molecules having only 2 lone pair of electrons $=3$
Which are $\mathrm{H}_2 \mathrm{O} , \mathrm{N}_2$ and $\mathrm{CO}$
Note:-
$\mathrm{XeF}_4$ have 2 lps on central atom, but we are asked about lone pair in molecule
The number of bent-shaped molecule/s from the following is __________
N$_3^-$, NO$_2^-$, I$_3^-$, O$_3$, SO$_2$
Explanation:
A bent-shaped molecule has a molecular geometry with a central atom bonded to two other atoms and one or two pairs of non-bonding electrons. The VSEPR (Valence Shell Electron Pair Repulsion) theory helps us predict the shapes of molecules.
Let's consider the given molecules:
N$_3^-$: The azide ion has a linear shape, not bent. The nitrogen in the middle is connected to two other nitrogens and there are no lone pairs on the central atom.
NO$_2^-$: The nitrite ion has a bent shape. The central nitrogen atom is connected to two oxygen atoms and has one lone pair of electrons, giving it a bent geometry.
I$_3^-$: The triiodide ion has a linear shape, not bent. The central iodine atom is connected to two other iodine atoms and there are no lone pairs on the central atom.
O$_3$: Ozone has a bent shape. The central oxygen atom is connected to two other oxygen atoms and has one lone pair of electrons, giving it a bent geometry.
SO$_2$: Sulfur dioxide has a bent shape. The central sulfur atom is connected to two oxygen atoms and has one lone pair of electrons, giving it a bent geometry.
Therefore, there are 3 bent-shaped molecules: NO$_2^-$, O$_3$, and SO$_2$.
The sum of lone pairs present on the central atom of the interhalogen IF$_5$ and IF$_7$ is _________
Explanation:
Interhalogen compounds are the substances that consist of two different halogens. The most common type of interhalogen compounds are binary, containing only two different elements.
In IF$_5$, iodine (I) is the central atom. Iodine has 7 valence electrons, 5 of which are used for bonding with the 5 fluorine (F) atoms. This leaves 2 electrons, or 1 lone pair on the iodine atom.
In IF$_7$, iodine (I) is the central atom again. Iodine has 7 valence electrons, all of which are used for bonding with the 7 fluorine (F) atoms. So, there are no lone pairs on the iodine atom.
Therefore, the sum of lone pairs present on the central atom of the interhalogens IF$_5$ and IF$_7$ is 1.
The number of species from the following carrying a single lone pair on central atom Xenon is ___________.
$\mathrm{XeF}_{5}^{+}, \mathrm{XeO}_{3}, \mathrm{XeO}_{2} \mathrm{~F}_{2}, \mathrm{XeF}_{5}^{-}, \mathrm{XeO}_{3} \mathrm{~F}_{2}, \mathrm{XeOF}_{4}, \mathrm{XeF}_{4}$
Explanation:
So, Answer is 4
The number of following factors which affect the percent covalent character of the ionic bond is _________
(A) Polarising power of cation
(B) Extent of distortion of anion
(C) Polarisability of the anion
(D) Polarising power of anion
Explanation:
The covalent character of an ionic bond is largely determined by the polarization of the ions involved in the bond. Polarization refers to the distortion of the electron cloud of an anion by a cation. This distortion leads to a shift in electron density towards the cation, thereby increasing the covalent character of the bond.
Here's a breakdown of the options:
(A) Polarising power of cation: The greater the polarising power of a cation, the greater the distortion of the anion, and therefore, the greater the covalent character of the bond. Thus, this statement is correct.
(B) Extent of distortion of anion: The more an anion is distorted, the more the electron density is shifted towards the cation, and the greater the covalent character of the bond. This statement is also correct.
(C) Polarisability of the anion: The greater the polarisability of an anion, the more it can be distorted, and therefore, the greater the covalent character of the bond. Thus, this statement is also correct.
(D) Polarising power of anion: This statement is incorrect. It is not the polarising power of the anion, but the polarisability of the anion and the polarising power of the cation that influence the covalent character of the bond.
Therefore, three of the factors listed (A, B, C) affect the percent covalent character of the ionic bond.
The number of species having a square planar shape from the following is __________.
$\mathrm{XeF}_{4}, \mathrm{SF}_{4}, \mathrm{SiF}_{4}, \mathrm{BF}_{4}^{-}, \mathrm{BrF}_{4}^{-},\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+},\left[\mathrm{FeCl}_{4}\right]^{2-},\left[\mathrm{PtCl}_{4}\right]^{2-}$
Explanation:
$\mathrm{SF}_4 \rightarrow$ See saw
$\mathrm{SiF}_4 \rightarrow$ Tetrahedral
$\mathrm{BF}_4^{-} \rightarrow$ Tetrahedral
$\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+} \rightarrow$ Square planar
$\left[\mathrm{FeCl}_4\right]^{2-} \rightarrow$ Tetrahedral
$\left[\mathrm{PtCl}_4\right]^{2-} \rightarrow$ Square planar
$\mathrm{BrF}_4^{-} \rightarrow$ Square planar
So, 4 square planer shape compounds are present.
In an ice crystal, each water molecule is hydrogen bonded to ____________ neighbouring molecules.
Explanation:
The number of species from the following which have square pyramidal structure is _________
$\mathrm{PF}_{5}, \mathrm{BrF}_{4}^{-}, \mathrm{IF}_{5}, \mathrm{BrF}_{5}, \mathrm{XeOF}_{4}, \mathrm{ICl}_{4}^{-}$
Explanation:
A square pyramidal structure has five bonds and one lone pair, making a total of six electron pairs around the central atom. The geometry of such a molecule can be analyzed using the VSEPR theory.
- $\mathrm{PF}_{5}$: 5 bond pairs and 0 lone pairs, so its geometry is trigonal bipyramidal, not square pyramidal.
- $\mathrm{BrF}_{4}^{-}$: 4 bond pairs and 2 lone pairs, so its geometry is square planar, not square pyramidal.
- $\mathrm{IF}_{5}$: 5 bond pairs and 1 lone pair, so its geometry is square pyramidal.
- $\mathrm{BrF}_{5}$: 5 bond pairs and 1 lone pair, so its geometry is square pyramidal.
- $\mathrm{XeOF}_{4}$: 5 bond pairs and 1 lone pair, so its geometry is square pyramidal.
- $\mathrm{ICl}_{4}^{-}$: 4 bond pairs and 2 lone pairs, so its geometry is square planar, not square pyramidal.
So the number of species from the given list that have a square pyramidal structure is 3.
$\mathrm{XeF}_{2}, \mathrm{I}_{3}^{+}, \mathrm{C}_{3} \mathrm{O}_{2}, \mathrm{I}_{3}^{-}, \mathrm{CO}_{2}, \mathrm{SO}_{2}, \mathrm{BeCl}_{2}$ and $\mathrm{BCl}_{2}^{\ominus}$
Explanation:
The number of molecules or ions from the following, which do not have odd number of electrons are _________.
(A) NO$_2$
(B) ICl$_4^ - $
(C) BrF$_3$
(D) ClO$_2$
(E) NO$_2^ + $
(F) NO
Explanation:
$ \begin{aligned} & \mathrm{NO}_2 \Rightarrow 23 e^{-} ; \\\\ & \mathrm{ICl}_4^{-} \Rightarrow 122 e^{-} ; \\\\ & \mathrm{BrF}_3 \Rightarrow 62 e^{-} ; \\\\ & \mathrm{ClO}_2 \Rightarrow 33 e^{-} ; \\\\ & \mathrm{NO}_2^{+} \Rightarrow 22 e^{-} ; \\\\ & \mathrm{NO} \Rightarrow 15 e^{-} \end{aligned} $
The total number of lone pairs of electrons on oxygen atoms of ozone is __________.
Explanation:
Total no, of lone pairs on oxygen atoms = 6
Which of the following pair of molecules contain odd electron molecule and an expanded octet molecule?
Number of lone pairs of electrons in the central atom of $\mathrm{SCl}_{2}, \mathrm{O}_{3}, \mathrm{ClF}_{3}$ and $\mathrm{SF}_{6}$, respectively, are :
Match List - I with List - II.
| List - I | List - II | ||
|---|---|---|---|
| (A) | $\psi_{\mathrm{MO}}=\psi_{\mathrm{A}}-\psi_{\mathrm{B}}$ | (I) | Dipole moment |
| (B) | $\mu=Q \times r$ | (II) | Bonding molecular orbital |
| (C) | $\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}$ | (III) | Anti-bonding molecular orbital |
| (D) | $\psi_{\mathrm{MO}}=\psi_{\mathrm{A}}+\psi_{\mathrm{B}}$ | (IV) | Bond order |
Choose the correct answer from the options given below :
Given below are two statements.
Statement I: $\mathrm{O}_{2}, \mathrm{Cu}^{2+}$, and $\mathrm{Fe}^{3+}$ are weakly attracted by magnetic field and are magnetized in the same direction as magnetic field.
Statement II: $\mathrm{NaCl}$ and $\mathrm{H}_{2} \mathrm{O}$ are weakly magnetized in opposite direction to magnetic field.
In the light of the above statements, choose the most appropriate answer from the options given below.
Arrange the following in increasing order of their covalent character.
A. $\mathrm{CaF}_{2}$
B. $\mathrm{CaCl}_{2}$
C. $\mathrm{CaBr}_{2}$
D. $\mathrm{CaI}_{2}$
Choose the correct answer from the options given below.
Match List-I with List-II :
| List I (Compound) |
List II (Shape) |
||
|---|---|---|---|
| (A) | BrF$_5$ | (I) | bent |
| (B) | [CrF$_6$]$^{3 - }$ | (II) | square pyramidal |
| (C) | O$_3$ | (III) | trigonal bipyramidal |
| (D) | PCl$_5$ | (IV) | octahedral |
Choose the correct answer from the options given below :
Match List I with List II:
| List I (molecule) |
List II (hybridization ; shape) |
||
|---|---|---|---|
| (A) | XeO$_3$ | (I) | sp$^3$d ; linear |
| (B) | XeF$_2$ | (II) | sp$^3$ ; pyramidal |
| (C) | XeOF$_4$ | (III) | sp$^3$d$^3$ ; distorted octahedral |
| (D) | XeF$_6$ | (IV) | sp$^3$d$^2$ ; square pyramidal |
Choose the correct answer from the options given below:
Consider the species CH4, NH$_4^ + $ and BH$_4^ - $. Choose the correct option with respect to the these species.
Number of lone pair(s) of electrons on central atom and the shape BrF3 molecule respectively, are
In the structure of SF4, the lone pair of electrons on S is in.
Identify the incorrect statement for PCl5 from the following.
The correct order of increasing intermolecular hydrogen bond strength is :
Based upon VSEPR theory, match the shape (geometry) of the molecules in List-I with the molecules in List-II and select the most appropriate option.
| List - I (Shape) |
List - II (Molecules) |
||
|---|---|---|---|
| (A) | T-shaped | (I) | XeF$_4$ |
| (B) | Trigonal planar | (II) | SF$_4$ |
| (C) | Square planar | (III) | ClF$_3$ |
| (D) | See-saw | (IV) | BF$_3$ |
Consider the ions/molecule
O$_2^ + $, O2, O$_2^ - $, O$_2^ {2-} $
For increasing bond order the correct option is :
Bonding in which of the following diatomic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron?
(A) NO
(B) N2
(C) O2
(D) C2
(E) B2
Choose the most appropriate answer from the options given below :
Number of electron deficient molecules among the following
PH3, B2H6, CCl4, NH3, LiH and BCl3 is
The correct order of bond orders of ${C_2}^{2 - }$, ${N_2}^{2 - }$ and ${O_2}^{2 - }$
is, respectivelyConsider, $\mathrm{PF}_{5}, \mathrm{BrF}_{5}, \mathrm{PCl}_{3}, \mathrm{SF}_{6},\left[\mathrm{ICl}_{4}\right]^{-}, \mathrm{ClF}_{3}$ and $\mathrm{IF}_{5}$.
Amongst the above molecule(s)/ion(s), the number of molecule(s)/ion(s) having $\mathrm{sp}^{3}\mathrm{~d}^{2}$ hybridisation is __________.
Explanation:
The number of paramagnetic species among the following is ___________.
$\mathrm{B}_{2}, \mathrm{Li}_{2}, \mathrm{C}_{2}, \mathrm{C}_{2}^{-}, \mathrm{O}_{2}^{2-}, \mathrm{O}_{2}^{+}$ and $\mathrm{He}_{2}^{+}$
Explanation:
And those species which have no unpaired electrons are called diamagnetic species.
B2 has 10 electrons.
Molecular orbital configuration of B2 is
${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$
Here two unpaired electrons present. So it is paramagnetic.
$O_2^{2−}$ has 18 electrons.
Moleculer orbital configuration of $O_2^{2−}$ is
${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $
Here is no unpaired electron so it is diamagnetic.
$O_2^{+}$ has 15 electrons.
Moleculer orbital configuration of $O_2^{+}$ is
${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $
Here 1 unpaired electron present, so it is paramagnetic.
$C_2$ has 12 electrons.
Moleculer orbital configuration of $C_2$
= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}$
Here no unpaired electron present, so it is diamagnetic.
$C_2^{ - }$ has 13 electrons.
Moleculer orbital configuration of $C_2^{ - }$ is
${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$
Here 1 unpaired electron present, so it is paramagnetic.
Li2 has 6 electrons.
Li2 = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}} \,$
Here no unpaired electron present, so it is diamagnetic.
Configuration of $He_2^ + $ (3 electrons) is = ${\sigma _{1{s^2}}}$ $\sigma _{1{s^1}}^ * $
Here 1 unpaired electron present, so it is paramagnetic.
The number of interhalogens from the following having square pyramidal structure is :
$\mathrm{ClF}_{3}, \mathrm{IF}_{7}, \mathrm{BrF}_{5}, \mathrm{BrF}_{3}, \mathrm{I}_{2} \mathrm{Cl}_{6}, \mathrm{IF}_{5}, \mathrm{ClF}, \mathrm{ClF}_{5}$
Explanation:
$\mathrm{IF}_{7} \rightarrow 7 \sigma$ bond $+0$ lone pair
$\mathrm{BrF}_{5} \rightarrow 5 \sigma$ bond $+1$ lone pair $\rightarrow$ Square pyramidal
$\mathrm{BrF}_{3} \rightarrow 3 \sigma$ bond $+2$ lone pair
$\mathrm{I}_{2} \mathrm{Cl}_{6} \rightarrow 4 \sigma$ bond $+2$ lone pair
$\mathrm{IF}_{5} \rightarrow 5 \sigma$ bond $+1$ lone pair $\rightarrow$ Square pyramidal
$\mathrm{CIF} \rightarrow 1 \sigma$ bond $+3$ lone pair
$\mathrm{CIF}_{5} \rightarrow 5 \sigma$ bond $+1$ lone pair $\rightarrow$ Square pyramidal
The number of molecule(s) or ion(s) from the following having non-planar structure is ____________.
$\mathrm{NO}_{3}^{-}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{BF}_{3}, \mathrm{PCl}_{3}, \mathrm{XeF}_{4}, \mathrm{SF}_{4}, \mathrm{XeO}_{3}, \mathrm{PH}_{4}^{+}, \mathrm{SO}_{3},\left[\mathrm{Al}(\mathrm{OH})_{4}\right]^{-}$
Explanation:
$\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow$ Open book (Non-planar)
$\mathrm{BF}_{3} \rightarrow$ Trigonal planar (Planar)
$\mathrm{PCl}_{3} \rightarrow$ Pyramidal (Non-planar)
$\mathrm{XeF}_{4} \rightarrow$ Square planar (Planar)
$\mathrm{SF}_{4} \rightarrow$ See-Saw (Non-planar) $\mathrm{XeO}_{3} \rightarrow$ Pyramidal (Non-planar)
$\mathrm{PH}_{4}^{\oplus} \rightarrow$ Tetrahedral (Non-planar)
$\mathrm{SO}_{3} \rightarrow$ Trigonal planar (Planar)
$\left[\mathrm{Al}(\mathrm{OH})_{4}\right]^{-} \rightarrow$ Tetrahedral (Non-planar)
Amongst the following, the number of oxide(s) which are paramagnetic in nature is
$\mathrm{Na}_{2} \mathrm{O}, \mathrm{KO}_{2}, \mathrm{NO}_{2}, \mathrm{~N}_{2} \mathrm{O}, \mathrm{ClO}_{2}, \mathrm{NO}, \mathrm{SO}_{2}, \mathrm{Cl}_{2} \mathrm{O}$
Explanation:
Diamagnetic species are: $\mathrm{Na}_{2} \mathrm{O}, \mathrm{N}_{2} \mathrm{O}, \mathrm{SO}_{2}, \mathrm{Cl}_{2} \mathrm{O}$
According to MO theory, number of species/ions from the following having identical bond order is ________.
$\mathrm{CN}^{-}, \mathrm{NO}^{+}, \mathrm{O}_{2}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2}^{2+}$
Explanation:
$\mathrm{O}_{2}$ has bond order of 2
$\mathrm{O}_{2}^{+}$ has bond order of $2.5$
$\therefore 3$ species have similar bond order.
The sum of number of lone pairs of electrons present on the central atoms of XeO3, XeOF4 and XeF6, is ______________
Explanation:
From structure, it is clear that it has five bond pairs and one lone pair.
Among the following species
$\mathrm{N}_{2}, \mathrm{~N}_{2}^{+}, \mathrm{N}_{2}^{-}, \mathrm{N}_{2}^{2-}, \mathrm{O}_{2}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{2-}$
the number of species showing diamagnesim is _______________.
Explanation:
And those species which have no unpaired electrons are called diamagnetic species.
(1) $N_2$ has 14 electrons.
Moleculer orbital configuration of $N_2$
= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$
Here no unpaired electron present, so it is diamagnetic.
(2) Moleculer orbital configuration of $N_2^{ + }$ (13 electrons)
= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$
Here in $N_2^{ + }$, 1 unpaired electron present, so it is paramagnetic.
(3) $\mathrm{N}_{2}^{2-}$ has 16 electrons.
Moleculer orbital configuration of $\mathrm{N}_{2}^{2-}$ is
${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $
Here 2 unpaired electron present, so it is paramagnetic.
(4) $\mathrm{N}_{2}^{-}$ has 15 electrons.
Moleculer orbital configuration of $\mathrm{N}_{2}^{-}$ is
${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $
Here 1 unpaired electron present, so it is paramagnetic.
(a) $O_2^{2−}$ has 18 electrons.
Moleculer orbital configuration of $O_2^{2−}$ is
${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $
Here is no unpaired electron so it is diamagnetic.
(b) $O_2^{−}$ has 17 electrons.
Moleculer orbital configuration of $O_2^{2−}$ is
${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $
Here 1 unpaired electron present, so it is paramagnetic.
(c) $O_2$ has 16 electrons.
Moleculer orbital configuration of $O_2$ is
${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $
Here 2 unpaired electron present, so it is paramagnetic.
(d) $O_2^{+}$ has 15 electrons.
Moleculer orbital configuration of $O_2^{+}$ is
${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $
Here 1 unpaired electron present, so it is paramagnetic.
Amongst the following, the number of molecule/(s) having net resultant dipole moment is ____________.
NF3, BF3, BeF2, CHCl3, H2S, SiF4, CCl4, PF5
Explanation:
Unsymmetrical molecules have net diploe moment like $-\mathrm{NF}_3$, $\mathrm{CHCl}_3$ and $\mathrm{H}_2 \mathrm{S}$
The hybridization of P exhibited in PF5 is spxdy. The value of y is __________.
Explanation:
(5 sigma bonds, zero lone pair on central atom)
Value of $y=1$
Amongst SF4, XeF4, CF4 and H2O, the number of species with two lone pairs of electrons is _____________.
Explanation:
Amongst BeF2, BF3, H2O, NH3, CCl4 and HCl, the number of molecules with non-zero net dipole moment is ____________.
Explanation:
$\mathrm{H}_2 \mathrm{O}, \mathrm{NH}_3$ and $\mathrm{HCl} \Rightarrow \mu_{\mathrm{net}} \neq 0$
Li2O, CaO, Na2O2, KO2, MgO and K2O
Choose the most appropriate answer from the options given below :
| List - I (Property) |
List - II (Example) |
||
|---|---|---|---|
| (a) | Diamagnetism | (i) | MnO |
| (b) | Ferrimagnetism | (ii) | ${O_2}$ |
| (c) | Paramagnetism | (iii) | NaCl |
| (d) | Antiferromagnetism | (iv) | $F{e_3}{O_4}$ |
Choose the most appropriate answer from the options given below :
| List-I (Species) |
List-II (Hybrid Orbitals) |
||
|---|---|---|---|
| (a) | $S{F_4}$ | (i) | $s{p^3}{d^2}$ |
| (b) | $I{F_5}$ | (ii) | ${d^2}s{p^3}$ |
| (c) | $NO_2^ + $ | (iii) | $s{p^3}d$ |
| (d) | $NH_4^ + $ | (iv) | $s{p^3}$ |
| (v) | $sp$ |
Choose the correct answer from the options given below :