ChemistryQ1–7 of 7 questions
Page 1 of 1 · English
1
ChemistryEasyMCQGUJCET · 2009
What is the atomic number of the element with $M^{2+}$ ion having electronic configuration $[Ar] 3d^8$?
A
$26$
B
$27$
C
$28$
D
$25$
Solution
(C) The electronic configuration of the $M^{2+}$ ion is $[Ar] 3d^8$.
This means the neutral atom $M$ has lost $2$ electrons to form the $M^{2+}$ ion.
To find the configuration of the neutral atom $M$,we add the $2$ electrons back to the $3d^8$ configuration.
The configuration of the neutral atom $M$ is $[Ar] 3d^8 4s^2$.
The total number of electrons in the neutral atom is $18$ (from $[Ar]$) $+ 8 + 2 = 28$.
Therefore,the atomic number of the element is $28$.
This means the neutral atom $M$ has lost $2$ electrons to form the $M^{2+}$ ion.
To find the configuration of the neutral atom $M$,we add the $2$ electrons back to the $3d^8$ configuration.
The configuration of the neutral atom $M$ is $[Ar] 3d^8 4s^2$.
The total number of electrons in the neutral atom is $18$ (from $[Ar]$) $+ 8 + 2 = 28$.
Therefore,the atomic number of the element is $28$.
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2
ChemistryEasyMCQGUJCET · 2009
What is the formula to find the value of $t_{1/2}$ for a Zero Order Reaction?
A
$\frac{0.693}{k}$
B
$\frac{[R]_0}{2k}$
C
$\frac{2k}{[R]_0}$
D
$\frac{k}{[R]_0}$
Solution
(B) For a zero order reaction,the integrated rate equation is $[R] = [R]_0 - kt$.
At half-life,$t = t_{1/2}$ and $[R] = \frac{[R]_0}{2}$.
Substituting these values into the equation: $\frac{[R]_0}{2} = [R]_0 - kt_{1/2}$.
Rearranging for $t_{1/2}$: $kt_{1/2} = [R]_0 - \frac{[R]_0}{2} = \frac{[R]_0}{2}$.
Therefore,$t_{1/2} = \frac{[R]_0}{2k}$.
The correct option is $B$.
At half-life,$t = t_{1/2}$ and $[R] = \frac{[R]_0}{2}$.
Substituting these values into the equation: $\frac{[R]_0}{2} = [R]_0 - kt_{1/2}$.
Rearranging for $t_{1/2}$: $kt_{1/2} = [R]_0 - \frac{[R]_0}{2} = \frac{[R]_0}{2}$.
Therefore,$t_{1/2} = \frac{[R]_0}{2k}$.
The correct option is $B$.
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3
ChemistryEasyMCQGUJCET · 2009
For a fourth-order reaction,what is the unit of $K$?
A
$(\text{mole} / \text{litre})^{-3}$
B
$(\text{mole} / \text{litre})^{-3} \ s$
C
$(\text{mole} / \text{litre})^{+3} \ s^{-1}$
D
$(\text{mole} / \text{litre})^{-3} \ s^{-1}$
Solution
(D) The general formula for the rate constant $K$ of an $n^{th}$ order reaction is given by:
$K = (\text{mole} / \text{litre})^{1-n} \cdot s^{-1}$
For a fourth-order reaction,$n = 4$.
Substituting $n = 4$ into the formula:
$K = (\text{mole} / \text{litre})^{1-4} \cdot s^{-1}$
$K = (\text{mole} / \text{litre})^{-3} \cdot s^{-1}$
Therefore,the correct option is $D$.
$K = (\text{mole} / \text{litre})^{1-n} \cdot s^{-1}$
For a fourth-order reaction,$n = 4$.
Substituting $n = 4$ into the formula:
$K = (\text{mole} / \text{litre})^{1-4} \cdot s^{-1}$
$K = (\text{mole} / \text{litre})^{-3} \cdot s^{-1}$
Therefore,the correct option is $D$.
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4
ChemistryEasyMCQGUJCET · 2009
How many electrons are present in the $3d$ orbital of the tetrahedral $K_2[NiCl_4]$ complex?
A
$10$ electrons
B
$8$ electrons
C
$6$ electrons
D
$7$ electrons
Solution
(B) In the complex $K_2[NiCl_4]$,let the oxidation state of $Ni$ be $x$.
$2(+1) + x + 4(-1) = 0$ $\Rightarrow 2 + x - 4 = 0$ $\Rightarrow x = +2$.
The atomic number of $Ni$ is $28$.
The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
Thus,the number of electrons in the $3d$ orbital is $8$.
$2(+1) + x + 4(-1) = 0$ $\Rightarrow 2 + x - 4 = 0$ $\Rightarrow x = +2$.
The atomic number of $Ni$ is $28$.
The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
Thus,the number of electrons in the $3d$ orbital is $8$.
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5
ChemistryEasyMCQGUJCET · 2009
The name of the ring structure complex compound formed between a metal ion and a polydentate ligand is . . . . . . .
A
Polynuclear complex
B
Chelate complex
C
Simple complex
D
None of these
Solution
(B) When a polydentate ligand binds to a metal ion through two or more donor atoms,it forms a cyclic or ring-like structure known as a chelate ring.
Such complexes are called $Chelate \ complexes$.
Therefore,the correct option is $B$.
Such complexes are called $Chelate \ complexes$.
Therefore,the correct option is $B$.
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6
ChemistryEasyMCQGUJCET · 2009
Which complex compound possesses $sp^3d^2$ hybridisation?
A
$[Fe(Cl)_6]^{3-}$
B
$[Fe(CN)_6]^{3-}$
C
$[Fe(CN)_6]^{4-}$
D
$[Fe(NH_3)_6]^{3+}$
Solution
(A) In $[Fe(Cl)_6]^{3-}$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Since $Cl^-$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the $3d$ electrons remain unpaired,and the complex utilizes the $4s$,$4p$,and $4d$ orbitals for hybridization,resulting in $sp^3d^2$ hybridization.
This is an outer orbital complex.
In contrast,$[Fe(CN)_6]^{3-}$,$[Fe(CN)_6]^{4-}$,and $[Fe(NH_3)_6]^{3+}$ involve strong field ligands (or specific electronic conditions) that lead to $d^2sp^3$ hybridization (inner orbital complexes).
Since $Cl^-$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the $3d$ electrons remain unpaired,and the complex utilizes the $4s$,$4p$,and $4d$ orbitals for hybridization,resulting in $sp^3d^2$ hybridization.
This is an outer orbital complex.
In contrast,$[Fe(CN)_6]^{3-}$,$[Fe(CN)_6]^{4-}$,and $[Fe(NH_3)_6]^{3+}$ involve strong field ligands (or specific electronic conditions) that lead to $d^2sp^3$ hybridization (inner orbital complexes).
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7
ChemistryEasyMCQGUJCET · 2009
Which Lanthanoid compound is used as a pigment?
A
$Tb(OH)_3$
B
$Ce(OH)_3$
C
$Lu(OH)_3$
D
$CeO_2$
Solution
(D) $CeO_2$ (Cerium dioxide) is widely used as a pigment in the glass and ceramic industries. It is also used as an opacifier and in the polishing of glass.
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