ChemistryQ1–29 of 29 questions
Page 1 of 1 · English
1
ChemistryEasyMCQGUJCET · 2019
If $P$ and $S$ are toluene,identify $Q$ and $R$ respectively in the following reaction sequence:
$P$ $\xrightarrow{KMnO_4/KOH} Q$ $\xrightarrow{\text{soda lime}, \Delta} R$ $\xrightarrow{CH_3Cl, \text{anhydrous } AlCl_3} S$
$P$ $\xrightarrow{KMnO_4/KOH} Q$ $\xrightarrow{\text{soda lime}, \Delta} R$ $\xrightarrow{CH_3Cl, \text{anhydrous } AlCl_3} S$
A
Benzoic acid,Benzene
B
Benzaldehyde,Benzoic Acid
C
Benzaldehyde,sodium benzoate
D
Benzene,Benzoic Acid
Solution
(A) $1$. $P$ is toluene $(C_6H_5CH_3)$.
$2$. Oxidation of toluene with $KMnO_4/KOH$ gives potassium benzoate,which upon acidification yields benzoic acid $(C_6H_5COOH)$. Thus,$Q$ is benzoic acid.
$3$. Decarboxylation of benzoic acid with soda lime $(NaOH + CaO)$ gives benzene $(C_6H_6)$. Thus,$R$ is benzene.
$4$. Friedel-Crafts alkylation of benzene with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ gives toluene $(S)$.
$5$. Therefore,$Q$ is benzoic acid and $R$ is benzene.
$2$. Oxidation of toluene with $KMnO_4/KOH$ gives potassium benzoate,which upon acidification yields benzoic acid $(C_6H_5COOH)$. Thus,$Q$ is benzoic acid.
$3$. Decarboxylation of benzoic acid with soda lime $(NaOH + CaO)$ gives benzene $(C_6H_6)$. Thus,$R$ is benzene.
$4$. Friedel-Crafts alkylation of benzene with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ gives toluene $(S)$.
$5$. Therefore,$Q$ is benzoic acid and $R$ is benzene.
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2
ChemistryEasyMCQGUJCET · 2019
$A$ polarimeter is used to determine . . . . . . of compounds.
A
$D$ and $L$ configuration
B
$d$ and $l$ configuration
C
$R$ and $S$ configuration
D
Both $D$ and $L$ as well as $d$ and $l$ configuration
Solution
(B) polarimeter is an instrument used to measure the angle of rotation caused by passing polarized light through an optically active substance.
This rotation determines whether a compound is dextrorotatory ($d$ or $+$) or levorotatory ($l$ or $-$).
Therefore,it is used to determine the $d$ and $l$ configuration (optical rotation) of compounds.
This rotation determines whether a compound is dextrorotatory ($d$ or $+$) or levorotatory ($l$ or $-$).
Therefore,it is used to determine the $d$ and $l$ configuration (optical rotation) of compounds.
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3
ChemistryEasyMCQGUJCET · 2019
Give the $IUPAC$ name of Methyl Salicylate.
A
Methyl $2$-hydroxybenzoate
B
Methoxybenzoic acid
C
$2$-Hydroxybenzoic acid
D
Methyl $3$-hydroxybenzoate
Solution
(A) Methyl salicylate is an ester formed from salicylic acid ($2$-hydroxybenzoic acid) and methanol. \\ The $IUPAC$ name for the ester of salicylic acid with methanol is Methyl $2$-hydroxybenzoate. \\ Therefore,the correct option is $A$.
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4
ChemistryEasyMCQGUJCET · 2019
Which of the following compounds reacts with ethylmagnesium bromide and also decolourizes bromine water solution?
A
$2-$vinylphenol
B
Methyl $2-$cyanophenylacetate
C
$2-$cyanoacetophenone
D
$2-$methoxyphenylacrylonitrile
Solution
(A) The compound must satisfy two conditions:
$1$. It must react with ethylmagnesium bromide $(C_2H_5MgBr)$,which is a Grignard reagent. This requires the presence of an acidic hydrogen (like in $-OH$,$-COOH$,etc.) or a carbonyl/cyano group.
$2$. It must decolourize bromine water,which indicates the presence of an unsaturated bond (alkene or alkyne).
Option $A$ is $2$-vinylphenol. It contains a phenolic $-OH$ group (acidic hydrogen) which reacts with $C_2H_5MgBr$ to form ethane gas. It also contains a vinyl group $(-CH=CH_2)$ which reacts with bromine water,causing decolourisation.
Therefore,the correct answer is $A$.
$1$. It must react with ethylmagnesium bromide $(C_2H_5MgBr)$,which is a Grignard reagent. This requires the presence of an acidic hydrogen (like in $-OH$,$-COOH$,etc.) or a carbonyl/cyano group.
$2$. It must decolourize bromine water,which indicates the presence of an unsaturated bond (alkene or alkyne).
Option $A$ is $2$-vinylphenol. It contains a phenolic $-OH$ group (acidic hydrogen) which reacts with $C_2H_5MgBr$ to form ethane gas. It also contains a vinyl group $(-CH=CH_2)$ which reacts with bromine water,causing decolourisation.
Therefore,the correct answer is $A$.
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5
ChemistryEasyMCQGUJCET · 2019
Which is the major product obtained by hydrolysis of the compound formed by the reaction between formaldehyde and ethyl magnesium bromide?
A
Propan-$1$-ol
B
Ethan-$1$-ol
C
Propan-$2$-ol
D
$2$-Methyl-Propan-$2$-ol
Solution
(A) The reaction between formaldehyde $(HCHO)$ and ethyl magnesium bromide $(CH_3CH_2MgBr)$ is a Grignard reaction.
Step $1$: Nucleophilic attack of the ethyl group on the carbonyl carbon of formaldehyde forms an intermediate alkoxide: $HCHO + CH_3CH_2MgBr \rightarrow CH_3CH_2CH_2OMgBr$.
Step $2$: Acidic hydrolysis of the intermediate yields the final alcohol: $CH_3CH_2CH_2OMgBr + H_2O \rightarrow CH_3CH_2CH_2OH + Mg(OH)Br$.
The product formed is $CH_3CH_2CH_2OH$,which is Propan-$1$-ol.
Therefore,the correct option is $A$.
Step $1$: Nucleophilic attack of the ethyl group on the carbonyl carbon of formaldehyde forms an intermediate alkoxide: $HCHO + CH_3CH_2MgBr \rightarrow CH_3CH_2CH_2OMgBr$.
Step $2$: Acidic hydrolysis of the intermediate yields the final alcohol: $CH_3CH_2CH_2OMgBr + H_2O \rightarrow CH_3CH_2CH_2OH + Mg(OH)Br$.
The product formed is $CH_3CH_2CH_2OH$,which is Propan-$1$-ol.
Therefore,the correct option is $A$.
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6
ChemistryEasyMCQGUJCET · 2019
Which of the following alcohols has the highest boiling point?
A
propan$-2-$ol
B
butan$-2-$ol
C
$2-$Methylpropan$-2-$ol
D
butan$-1-$ol
Solution
(D) The boiling point of alcohols depends on the extent of hydrogen bonding and the surface area of the molecule.
$1$. Among the given options,$butan-1-ol$ is a primary alcohol with a straight-chain structure,which provides a larger surface area for van der Waals forces compared to branched isomers.
$2$. $Propan-2-ol$ has $3$ carbons,$butan-2-ol$ is a secondary alcohol,and $2-Methylpropan-2-ol$ is a tertiary alcohol.
$3$. As branching increases,the surface area decreases,leading to weaker intermolecular forces and lower boiling points.
$4$. Therefore,$butan-1-ol$ has the highest boiling point among the given options due to its linear structure and higher molecular mass compared to $propan-2-ol$.
$1$. Among the given options,$butan-1-ol$ is a primary alcohol with a straight-chain structure,which provides a larger surface area for van der Waals forces compared to branched isomers.
$2$. $Propan-2-ol$ has $3$ carbons,$butan-2-ol$ is a secondary alcohol,and $2-Methylpropan-2-ol$ is a tertiary alcohol.
$3$. As branching increases,the surface area decreases,leading to weaker intermolecular forces and lower boiling points.
$4$. Therefore,$butan-1-ol$ has the highest boiling point among the given options due to its linear structure and higher molecular mass compared to $propan-2-ol$.
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7
ChemistryEasyMCQGUJCET · 2019
What is the formula of Acrolein?
A
$CH_2=CH-CHO$
B
$CH_2=CH-CN$
C
$CH_2=CH-COOH$
D
$CH_2=CH-CONH_2$
Solution
(A) Acrolein is the simplest unsaturated aldehyde. Its chemical structure consists of a vinyl group attached to an aldehyde group. The chemical formula is $CH_2=CH-CHO$. Therefore,the correct option is $A$.
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8
ChemistryEasyMCQGUJCET · 2019
Which of the following compounds undergoes aldol condensation?
A
Trimethyl acetaldehyde
B
Formaldehyde
C
Trichloro acetaldehyde
D
Acetaldehyde
Solution
(D) Aldol condensation occurs in aldehydes or ketones that possess at least one $\alpha$-hydrogen atom.
$A$. Trimethyl acetaldehyde $(CH_3)_3CCHO$ has no $\alpha$-hydrogen.
$B$. Formaldehyde $(HCHO)$ has no $\alpha$-hydrogen.
$C$. Trichloro acetaldehyde $(CCl_3CHO)$ has no $\alpha$-hydrogen.
$D$. Acetaldehyde $(CH_3CHO)$ has three $\alpha$-hydrogen atoms attached to the $\alpha$-carbon.
Therefore,acetaldehyde undergoes aldol condensation.
$A$. Trimethyl acetaldehyde $(CH_3)_3CCHO$ has no $\alpha$-hydrogen.
$B$. Formaldehyde $(HCHO)$ has no $\alpha$-hydrogen.
$C$. Trichloro acetaldehyde $(CCl_3CHO)$ has no $\alpha$-hydrogen.
$D$. Acetaldehyde $(CH_3CHO)$ has three $\alpha$-hydrogen atoms attached to the $\alpha$-carbon.
Therefore,acetaldehyde undergoes aldol condensation.
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9
ChemistryEasyMCQGUJCET · 2019
$C_6H_5COCl + C_6H_5COONa \xrightarrow{\Delta} \text{ . . . . . . }$.
A
Benzyl benzoate
B
Benzaldehyde
C
Benzyl Alcohol
D
Benzoic anhydride
Solution
(D) The reaction between benzoyl chloride $(C_6H_5COCl)$ and sodium benzoate $(C_6H_5COONa)$ is a nucleophilic acyl substitution reaction.
In this reaction,the carboxylate oxygen of the sodium benzoate acts as a nucleophile and attacks the carbonyl carbon of the benzoyl chloride,displacing the chloride ion.
This results in the formation of benzoic anhydride $((C_6H_5CO)_2O)$ and sodium chloride $(NaCl)$ as a byproduct.
The reaction is represented as: $C_6H_5COCl + C_6H_5COONa \xrightarrow{\Delta} (C_6H_5CO)_2O + NaCl$.
Therefore,the correct option is $D$.
In this reaction,the carboxylate oxygen of the sodium benzoate acts as a nucleophile and attacks the carbonyl carbon of the benzoyl chloride,displacing the chloride ion.
This results in the formation of benzoic anhydride $((C_6H_5CO)_2O)$ and sodium chloride $(NaCl)$ as a byproduct.
The reaction is represented as: $C_6H_5COCl + C_6H_5COONa \xrightarrow{\Delta} (C_6H_5CO)_2O + NaCl$.
Therefore,the correct option is $D$.
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10
ChemistryEasyMCQGUJCET · 2019
Assign '$T$' for true statements and '$F$' for false statements,and select the correct option from the given choices for the following statements:
$(i)$ Cytosine base is a derivative of pyrimidine.
$(ii)$ $\beta-D$-ribose sugar is present in $DNA$.
$(iii)$ The message for the synthesis of a specific protein is present in $RNA$.
$(iv)$ $DNA$ is responsible for maintaining the identity of different species of organisms for millions of years.
$(i)$ Cytosine base is a derivative of pyrimidine.
$(ii)$ $\beta-D$-ribose sugar is present in $DNA$.
$(iii)$ The message for the synthesis of a specific protein is present in $RNA$.
$(iv)$ $DNA$ is responsible for maintaining the identity of different species of organisms for millions of years.
A
$FTFF$
B
$TFFT$
C
$FFFT$
D
$FFTF$
Solution
(B) $(i)$ Cytosine is a pyrimidine derivative,so it is $T$.
$(ii)$ $DNA$ contains $2$-deoxy-$\beta-D$-ribose,not $\beta-D$-ribose (which is in $RNA$),so it is $F$.
$(iii)$ $RNA$ carries the genetic message for protein synthesis,so it is $T$.
$(iv)$ $DNA$ is responsible for the transmission of hereditary information over generations,not just for one century,so it is $F$.
Thus,the sequence is $T, F, T, F$. The correct option is $B$.
$(ii)$ $DNA$ contains $2$-deoxy-$\beta-D$-ribose,not $\beta-D$-ribose (which is in $RNA$),so it is $F$.
$(iii)$ $RNA$ carries the genetic message for protein synthesis,so it is $T$.
$(iv)$ $DNA$ is responsible for the transmission of hereditary information over generations,not just for one century,so it is $F$.
Thus,the sequence is $T, F, T, F$. The correct option is $B$.
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11
ChemistryEasyMCQGUJCET · 2019
Which protein present in muscles is insoluble in water?
A
Albumin
B
Carotene
C
Insulin
D
Myosin
Solution
(D) Proteins are classified into two types based on their molecular shape: fibrous and globular.
Fibrous proteins are long,thread-like structures that are generally insoluble in water.
$Myosin$ is a fibrous protein found in muscles,which makes it insoluble in water.
In contrast,$Albumin$ and $Insulin$ are globular proteins,which are generally soluble in water.
Therefore,the correct option is $D$.
Fibrous proteins are long,thread-like structures that are generally insoluble in water.
$Myosin$ is a fibrous protein found in muscles,which makes it insoluble in water.
In contrast,$Albumin$ and $Insulin$ are globular proteins,which are generally soluble in water.
Therefore,the correct option is $D$.
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12
ChemistryEasyMCQGUJCET · 2019
Why is glucose called gluco-pyranose?
A
Glucose is a ketohexose.
B
Glucose is an aldohexose.
C
Glucose is a cyclic compound containing $5$ carbon atoms and $1$ oxygen atom in the ring.
D
Glucose is a cyclic compound containing $6$ carbon atoms.
Solution
(C) The cyclic structure of glucose is called glucopyranose because it resembles the structure of the heterocyclic compound pyran,which contains a six-membered ring consisting of $5$ carbon atoms and $1$ oxygen atom. Therefore,the correct option is $C$.
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13
ChemistryEasyMCQGUJCET · 2019
Which statement is incorrect for collision theory?
A
The reactant experiencing fruitful collisions are converted to products.
B
There must be a certain minimum energy for the reactant experiencing collision.
C
The collision of the reactant molecules should be from any direction.
D
The collision between the reacting molecules is essential.
Solution
(C) According to collision theory,for a reaction to occur,molecules must collide with both sufficient kinetic energy (activation energy) and the correct orientation (steric factor). Option $C$ is incorrect because collisions must occur with a specific,favorable orientation to be effective; they cannot be from any random direction.
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14
ChemistryEasyMCQGUJCET · 2019
In a reaction $A \rightarrow B$,if the concentration of reactant is increased by $9$ times,then the rate of reaction increases $3$ times. What is the order of reaction?
A
$1/3$
B
$1/2$
C
$3$
D
$2$
Solution
(B) For a reaction $A \rightarrow B$,the rate law is given by: $\text{Rate} = k[A]^n$,where $n$ is the order of reaction.
Given that when the concentration $[A]$ is increased by $9$ times,the rate increases by $3$ times.
So,$3 \times \text{Rate} = k(9[A])^n$.
Dividing this by the original rate equation: $\frac{3 \times \text{Rate}}{\text{Rate}} = \frac{k(9[A])^n}{k[A]^n}$.
$3 = 9^n$.
Since $9 = 3^2$,we have $3 = (3^2)^n = 3^{2n}$.
Comparing the exponents: $1 = 2n$,which gives $n = 1/2$.
Therefore,the order of reaction is $1/2$.
Given that when the concentration $[A]$ is increased by $9$ times,the rate increases by $3$ times.
So,$3 \times \text{Rate} = k(9[A])^n$.
Dividing this by the original rate equation: $\frac{3 \times \text{Rate}}{\text{Rate}} = \frac{k(9[A])^n}{k[A]^n}$.
$3 = 9^n$.
Since $9 = 3^2$,we have $3 = (3^2)^n = 3^{2n}$.
Comparing the exponents: $1 = 2n$,which gives $n = 1/2$.
Therefore,the order of reaction is $1/2$.
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15
ChemistryEasyMCQGUJCET · 2019
The instantaneous rate of reaction for the reaction $3A + 2B \rightarrow 5C$ is . . . . . . .
A
$+\frac{1}{3} \frac{d[A]}{dt} = -\frac{1}{2} \frac{d[B]}{dt} = -\frac{1}{5} \frac{d[C]}{dt}$
B
$-\frac{1}{3} \frac{d[A]}{dt} = +\frac{1}{2} \frac{d[B]}{dt} = -\frac{1}{5} \frac{d[C]}{dt}$
C
$-\frac{1}{3} \frac{d[A]}{dt} = -\frac{1}{2} \frac{d[B]}{dt} = +\frac{1}{5} \frac{d[C]}{dt}$
D
$+\frac{1}{3} \frac{d[A]}{dt} = -\frac{1}{2} \frac{d[B]}{dt} = +\frac{1}{5} \frac{d[C]}{dt}$
Solution
(C) For a general reaction $aA + bB \rightarrow cC$,the rate of reaction is given by:
Rate $= -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = +\frac{1}{c} \frac{d[C]}{dt}$.
For the given reaction $3A + 2B \rightarrow 5C$,the coefficients are $a=3$,$b=2$,and $c=5$.
Substituting these values,we get:
Rate $= -\frac{1}{3} \frac{d[A]}{dt} = -\frac{1}{2} \frac{d[B]}{dt} = +\frac{1}{5} \frac{d[C]}{dt}$.
Therefore,option $C$ is correct.
Rate $= -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = +\frac{1}{c} \frac{d[C]}{dt}$.
For the given reaction $3A + 2B \rightarrow 5C$,the coefficients are $a=3$,$b=2$,and $c=5$.
Substituting these values,we get:
Rate $= -\frac{1}{3} \frac{d[A]}{dt} = -\frac{1}{2} \frac{d[B]}{dt} = +\frac{1}{5} \frac{d[C]}{dt}$.
Therefore,option $C$ is correct.
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16
ChemistryEasyMCQGUJCET · 2019
Which of the following complexes possesses a meridional $(mer)$ isomer?
A
$[Co(NH_3)_3Cl_3]$
B
$[Co(NH_3)_4Cl_2]$
C
$[Co(NH_3)_2Cl_4]$
D
$[Co(NH_3)_5Cl]$
Solution
(A) The meridional $(mer)$ isomer is a type of geometric isomerism found in octahedral complexes of the type $[MA_3B_3]$.
In this configuration,three identical ligands occupy the positions around the meridian of the octahedron.
Among the given options,$[Co(NH_3)_3Cl_3]$ is an octahedral complex of the type $[MA_3B_3]$.
It can exist in two geometric isomeric forms: facial $(fac)$ and meridional $(mer)$.
Therefore,the correct option is $A$.
In this configuration,three identical ligands occupy the positions around the meridian of the octahedron.
Among the given options,$[Co(NH_3)_3Cl_3]$ is an octahedral complex of the type $[MA_3B_3]$.
It can exist in two geometric isomeric forms: facial $(fac)$ and meridional $(mer)$.
Therefore,the correct option is $A$.
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17
ChemistryEasyMCQGUJCET · 2019
The primary valency and secondary valency of the central metal ion and the number of total ions produced in an aqueous solution for the $K[Co(OX)_2(NH_3)_2]$ complex are,respectively, . . . . . . .
A
$3, 4, 2$
B
$4, 4, 2$
C
$3, 6, 2$
D
$3, 6, 1$
Solution
(C) The complex is $K[Co(OX)_2(NH_3)_2]$.
$1$. Primary valency is the oxidation state of the central metal ion $(Co)$. Let the oxidation state be $x$. $1 + x + 2(-2) + 2(0) = 0$,so $x - 3 = 0$,which gives $x = +3$.
$2$. Secondary valency is the coordination number. $OX$ is a bidentate ligand (coordination number $2$) and $NH_3$ is a monodentate ligand (coordination number $1$). Total coordination number = $(2 \times 2) + (2 \times 1) = 4 + 2 = 6$.
$3$. In aqueous solution,the complex dissociates as: $K[Co(OX)_2(NH_3)_2] \rightarrow K^+ + [Co(OX)_2(NH_3)_2]^-$. This produces $1 + 1 = 2$ ions.
Thus,the values are $3, 6, 2$.
$1$. Primary valency is the oxidation state of the central metal ion $(Co)$. Let the oxidation state be $x$. $1 + x + 2(-2) + 2(0) = 0$,so $x - 3 = 0$,which gives $x = +3$.
$2$. Secondary valency is the coordination number. $OX$ is a bidentate ligand (coordination number $2$) and $NH_3$ is a monodentate ligand (coordination number $1$). Total coordination number = $(2 \times 2) + (2 \times 1) = 4 + 2 = 6$.
$3$. In aqueous solution,the complex dissociates as: $K[Co(OX)_2(NH_3)_2] \rightarrow K^+ + [Co(OX)_2(NH_3)_2]^-$. This produces $1 + 1 = 2$ ions.
Thus,the values are $3, 6, 2$.
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18
ChemistryEasyMCQGUJCET · 2019
Which of the following complex ions is the most stable?
A
$[Co(NH_3)_6]^{3+}$
B
$[CoCl_6]^{3-}$
C
$[CoF_6]^{3-}$
D
$[Co(H_2O)_6]^{3+}$
Solution
(A) The stability of a coordination complex depends on the nature of the ligand attached to the central metal ion.
$NH_3$ is a strong field ligand compared to $H_2O$,$F^-$,and $Cl^-$.
According to the spectrochemical series,the order of field strength is $Cl^- < F^- < H_2O < NH_3$.
Stronger ligands form more stable complexes with the central metal ion due to greater crystal field splitting energy.
Therefore,$[Co(NH_3)_6]^{3+}$ is the most stable complex among the given options.
$NH_3$ is a strong field ligand compared to $H_2O$,$F^-$,and $Cl^-$.
According to the spectrochemical series,the order of field strength is $Cl^- < F^- < H_2O < NH_3$.
Stronger ligands form more stable complexes with the central metal ion due to greater crystal field splitting energy.
Therefore,$[Co(NH_3)_6]^{3+}$ is the most stable complex among the given options.
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19
ChemistryEasyMCQGUJCET · 2019
What is the correct order for energy of $d$ orbitals during splitting in the $[NiCl_4]^{2-}$ complex ion?
A
$d_{xy} \cong d_{yz} \cong d_{xz} < d_{x^2-y^2} \cong d_{z^2}$
B
$d_{xy} \cong d_{yz} \cong d_{xz} \cong d_{x^2-y^2} \cong d_{z^2}$
C
$d_{x^2-y^2} \cong d_{z^2} < d_{xy} \cong d_{yz} \cong d_{xz}$
D
$d_{x^2-y^2} > d_{z^2} > d_{xy} \cong d_{yz} \cong d_{xz}$
Solution
(C) The complex ion $[NiCl_4]^{2-}$ is a tetrahedral complex.
In a tetrahedral crystal field,the $d$ orbitals split into two sets: a lower energy set $(d_{xy}, d_{yz}, d_{xz})$ and a higher energy set $(d_{x^2-y^2}, d_{z^2})$.
However,the splitting pattern for tetrahedral complexes is the inverse of octahedral complexes.
In tetrahedral geometry,the $d_{xy}, d_{yz}, d_{xz}$ orbitals (often denoted as $t_2$) are higher in energy than the $d_{x^2-y^2}, d_{z^2}$ orbitals (often denoted as $e$).
Therefore,the correct order of energy is $d_{x^2-y^2} \cong d_{z^2} < d_{xy} \cong d_{yz} \cong d_{xz}$.
In a tetrahedral crystal field,the $d$ orbitals split into two sets: a lower energy set $(d_{xy}, d_{yz}, d_{xz})$ and a higher energy set $(d_{x^2-y^2}, d_{z^2})$.
However,the splitting pattern for tetrahedral complexes is the inverse of octahedral complexes.
In tetrahedral geometry,the $d_{xy}, d_{yz}, d_{xz}$ orbitals (often denoted as $t_2$) are higher in energy than the $d_{x^2-y^2}, d_{z^2}$ orbitals (often denoted as $e$).
Therefore,the correct order of energy is $d_{x^2-y^2} \cong d_{z^2} < d_{xy} \cong d_{yz} \cong d_{xz}$.
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20
ChemistryEasyMCQGUJCET · 2019
Which of the following pairs has a similar magnetic moment?
A
$Cr^{3+}, Mn^{3+}$
B
$Fe^{3+}, Mn^{2+}$
C
$Fe^{2+}, Mn^{2+}$
D
$Ni^{2+}, Co^{2+}$
Solution
(B) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $Fe^{3+}$ $(Z=26)$: The electronic configuration is $[Ar] 3d^5$. The number of unpaired electrons $(n)$ = $5$.
For $Mn^{2+}$ $(Z=25)$: The electronic configuration is $[Ar] 3d^5$. The number of unpaired electrons $(n)$ = $5$.
Since both $Fe^{3+}$ and $Mn^{2+}$ have $5$ unpaired electrons,they have the same magnetic moment of $\sqrt{5(5+2)} = \sqrt{35} \ BM$.
For $Fe^{3+}$ $(Z=26)$: The electronic configuration is $[Ar] 3d^5$. The number of unpaired electrons $(n)$ = $5$.
For $Mn^{2+}$ $(Z=25)$: The electronic configuration is $[Ar] 3d^5$. The number of unpaired electrons $(n)$ = $5$.
Since both $Fe^{3+}$ and $Mn^{2+}$ have $5$ unpaired electrons,they have the same magnetic moment of $\sqrt{5(5+2)} = \sqrt{35} \ BM$.
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21
ChemistryEasyMCQGUJCET · 2019
Elements $A$ and $B$ do not form an alloy because....
A
Both elements have similar crystal structures
B
Radius of $A$ is $115 \ pm$ while radius of $B$ is $187 \ pm$
C
Both are the members of same group
D
Both have similar electronic configuration in valence shell
Solution
(B) The formation of an alloy requires the atomic radii of the two elements to differ by no more than $15\%$. In this case, the difference in radii is $(187 - 115) = 72 \ pm$. The percentage difference is $\frac{72}{115} \times 100 \approx 62.6\%$. Since this difference is significantly greater than $15\%$, the elements cannot form an alloy. Therefore, option $B$ is the correct reason.
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22
ChemistryEasyMCQGUJCET · 2019
At which temperature,ceramic materials behave as superconductors (in $K$)?
A
$90$
B
$15$
C
$200$
D
$150$
Solution
(D) Ceramic materials,specifically high-temperature superconductors like $YBa_2Cu_3O_7$,exhibit superconductivity at temperatures significantly higher than conventional metallic superconductors.
While many ceramic superconductors operate in the range of $90 \ K$ to $150 \ K$,the specific context of this question refers to the discovery of materials that show superconductivity at temperatures around $150 \ K$.
Therefore,the correct option is $D$.
While many ceramic superconductors operate in the range of $90 \ K$ to $150 \ K$,the specific context of this question refers to the discovery of materials that show superconductivity at temperatures around $150 \ K$.
Therefore,the correct option is $D$.
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23
ChemistryEasyMCQGUJCET · 2019
If $1$ mole of electrons is passed through the solutions of $AlCl_3$,$AgNO_3$,and $MgSO_4$,in what ratio will $Al$,$Ag$,and $Mg$ be deposited at the electrodes?
A
$3: 6: 2$
B
$2: 6: 3$
C
$1: 2: 3$
D
$3: 2: 1$
Solution
(B) The reduction reactions at the electrodes are as follows:
$Al^{3+} + 3e^- \rightarrow Al$
$Ag^+ + 1e^- \rightarrow Ag$
$Mg^{2+} + 2e^- \rightarrow Mg$
According to Faraday's laws of electrolysis,the number of moles of metal deposited is given by $n = \frac{\text{moles of electrons}}{\text{n-factor}}$.
For $Al$: $n_{Al} = \frac{1}{3} \text{ mole}$.
For $Ag$: $n_{Ag} = \frac{1}{1} = 1 \text{ mole}$.
For $Mg$: $n_{Mg} = \frac{1}{2} \text{ mole}$.
The ratio of moles deposited is $\frac{1}{3} : 1 : \frac{1}{2}$.
Multiplying by $6$ to simplify the ratio: $2 : 6 : 3$.
Thus,the correct option is $B$.
$Al^{3+} + 3e^- \rightarrow Al$
$Ag^+ + 1e^- \rightarrow Ag$
$Mg^{2+} + 2e^- \rightarrow Mg$
According to Faraday's laws of electrolysis,the number of moles of metal deposited is given by $n = \frac{\text{moles of electrons}}{\text{n-factor}}$.
For $Al$: $n_{Al} = \frac{1}{3} \text{ mole}$.
For $Ag$: $n_{Ag} = \frac{1}{1} = 1 \text{ mole}$.
For $Mg$: $n_{Mg} = \frac{1}{2} \text{ mole}$.
The ratio of moles deposited is $\frac{1}{3} : 1 : \frac{1}{2}$.
Multiplying by $6$ to simplify the ratio: $2 : 6 : 3$.
Thus,the correct option is $B$.
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24
ChemistryEasyMCQGUJCET · 2019
$Zn_{(s)} | Zn_{(aq)}^{2+}(1 \ M) || Ni_{(aq)}^{2+}(1 \ M) | Ni_{(s)}$
Which is incorrect for the given cell?
Which is incorrect for the given cell?
A
Electrochemical cell
B
Voltaic cell
C
Galvanic cell
D
Daniel cell
Solution
(D) The given cell is $Zn_{(s)} | Zn_{(aq)}^{2+}(1 \ M) || Ni_{(aq)}^{2+}(1 \ M) | Ni_{(s)}$.
This is an electrochemical cell,also known as a voltaic or galvanic cell,where chemical energy is converted into electrical energy.
$A$ $Daniel$ cell is a specific type of galvanic cell that uses a $Zn|Zn^{2+}$ anode and a $Cu|Cu^{2+}$ cathode.
Since this cell uses a $Ni|Ni^{2+}$ cathode instead of $Cu|Cu^{2+}$,it is not a $Daniel$ cell.
Therefore,the incorrect statement is $D$.
This is an electrochemical cell,also known as a voltaic or galvanic cell,where chemical energy is converted into electrical energy.
$A$ $Daniel$ cell is a specific type of galvanic cell that uses a $Zn|Zn^{2+}$ anode and a $Cu|Cu^{2+}$ cathode.
Since this cell uses a $Ni|Ni^{2+}$ cathode instead of $Cu|Cu^{2+}$,it is not a $Daniel$ cell.
Therefore,the incorrect statement is $D$.
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25
ChemistryEasyMCQGUJCET · 2019
Which one of the following groups of compounds acts as an extinguisher,antiseptic,insecticide,and anesthetic,respectively?
A
$CCl_4, CHI_3, DDT, CHCl_3$
B
$CHCl_3, CHI_3, DDT, CCl_4$
C
$DDT, CHCl_3, CCl_4, CHI_3$
D
$CCl_4, CHI_3, CHCl_3, DDT$
Solution
(A) $CCl_4$ (Carbon tetrachloride) is used as a fire extinguisher.
$CHI_3$ (Iodoform) is used as an antiseptic.
$DDT$ (Dichlorodiphenyltrichloroethane) is used as an insecticide.
$CHCl_3$ (Chloroform) is used as an anesthetic.
Therefore,the correct sequence is $CCl_4, CHI_3, DDT, CHCl_3$.
$CHI_3$ (Iodoform) is used as an antiseptic.
$DDT$ (Dichlorodiphenyltrichloroethane) is used as an insecticide.
$CHCl_3$ (Chloroform) is used as an anesthetic.
Therefore,the correct sequence is $CCl_4, CHI_3, DDT, CHCl_3$.
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26
ChemistryEasyMCQGUJCET · 2019
$1, 2-$Dichloroethane is which type of halide?
A
Alkylidene halide
B
Geminal halide
C
Vicinal halide
D
Allylic halide
Solution
(C) In $1, 2-$dichloroethane $(Cl-CH_2-CH_2-Cl)$,the two chlorine atoms are attached to adjacent carbon atoms.
Compounds in which two halogen atoms are attached to adjacent carbon atoms are known as vicinal dihalides or vicinal halides.
Therefore,the correct option is $C$.
Compounds in which two halogen atoms are attached to adjacent carbon atoms are known as vicinal dihalides or vicinal halides.
Therefore,the correct option is $C$.
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27
ChemistryEasyMCQGUJCET · 2019
The value of which of the following units of concentration will not change with the change in temperature?
A
Molarity
B
Molality
C
Normality
D
Formality
Solution
(B) Concentration units that involve volume (such as Molarity,Normality,and Formality) are temperature-dependent because volume changes with temperature.
Molality is defined as the number of moles of solute per kilogram of solvent.
Since mass does not change with temperature,the value of Molality remains constant regardless of temperature changes.
Therefore,the correct option is $B$.
Molality is defined as the number of moles of solute per kilogram of solvent.
Since mass does not change with temperature,the value of Molality remains constant regardless of temperature changes.
Therefore,the correct option is $B$.
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28
ChemistryEasyMCQGUJCET · 2019
If molality of a solution is $0.05 \ m$ and elevation in boiling point is $0.16 \ K$,then what is the molal elevation constant of the solvent?
A
$3.2 \ K \ kg \ mol^{-1}$
B
$1.6 \ K \ kg \ mol^{-1}$
C
$2.2 \ K \ kg \ mol^{-1}$
D
$2.3 \ K \ kg \ mol^{-1}$
Solution
(A) The formula for elevation in boiling point is given by $\Delta T_b = K_b \times m$,where $\Delta T_b$ is the elevation in boiling point,$K_b$ is the molal elevation constant,and $m$ is the molality of the solution.
Given values are $\Delta T_b = 0.16 \ K$ and $m = 0.05 \ m$.
Substituting these values into the formula: $0.16 = K_b \times 0.05$.
Solving for $K_b$: $K_b = \frac{0.16}{0.05} = 3.2 \ K \ kg \ mol^{-1}$.
Therefore,the correct option is $A$.
Given values are $\Delta T_b = 0.16 \ K$ and $m = 0.05 \ m$.
Substituting these values into the formula: $0.16 = K_b \times 0.05$.
Solving for $K_b$: $K_b = \frac{0.16}{0.05} = 3.2 \ K \ kg \ mol^{-1}$.
Therefore,the correct option is $A$.
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29
ChemistryEasyMCQGUJCET · 2019
Calculate Van't Hoff factor $(i)$ for an aqueous solution of $K_3[Fe(CN)_6]$ having a degree of dissociation $(\alpha)$ equal to $0.778$.
A
$4.334$
B
$3.334$
C
$0.222$
D
$2.334$
Solution
(B) The dissociation of $K_3[Fe(CN)_6]$ is given by:
$K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-}$
Here,the number of ions produced per formula unit $(n)$ is $3 + 1 = 4$.
The relationship between the Van't Hoff factor $(i)$,degree of dissociation $(\alpha)$,and the number of ions $(n)$ is:
$i = 1 + \alpha(n - 1)$
Given $\alpha = 0.778$ and $n = 4$:
$i = 1 + 0.778(4 - 1)$
$i = 1 + 0.778(3)$
$i = 1 + 2.334$
$i = 3.334$
Therefore,the correct option is $B$.
$K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-}$
Here,the number of ions produced per formula unit $(n)$ is $3 + 1 = 4$.
The relationship between the Van't Hoff factor $(i)$,degree of dissociation $(\alpha)$,and the number of ions $(n)$ is:
$i = 1 + \alpha(n - 1)$
Given $\alpha = 0.778$ and $n = 4$:
$i = 1 + 0.778(4 - 1)$
$i = 1 + 0.778(3)$
$i = 1 + 2.334$
$i = 3.334$
Therefore,the correct option is $B$.
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