Read the passage given below and answer the following questions: The molecular compounds which are formed from the combination of two or more simple stable compounds and retain their identity in the solid as well as in the dissolved state are called coordination compounds. Their properties are completely different from the constituents. ln coordination compounds, the central metal atom or ion is linked to a number ofions or neutral molecules, called ligands, by coordinate bonds. For example, Dimethylglyoxime (dmg) is a bid en date ligand chelating large amounts of metals. When dimethyl glyoxime is added to alcoholic solution of NiCl_2 and ammonium hydroxide is slowly added to it, a rosy red precipitate of a complex is formed. The following questions are multiple choice questions. Choose the most appropriate answer: 1. The structure of the complex is: 2. Oxidation number of Ni in the given complex is: 1. +3 2. +1 3. +2 4. Zero 3. Hybridisation and structure of the complex is: 1. Sp^3 , tetrahedral. 2. dsp^2 , square planar. 3. Sp^3 , square planar. 4. Sp^3d, trigonal bipyramidal. 4. Which of the following is true about this complex? 1. It is paramagnetic, containing 2 unpaired electrons. 2. It is paramagnetic, containing 1 unpaired electron. 3. It is paramagnetic, containing 4 unpaired electrons. 4. It is diamagnetic with no unpaired electron. 5. Which one will give test for Fe^ 3+ ions in the solution? 1. [Fe(CN)_6]^ 3- 2. [Fe(CN)_6]^ 2- 3. (NH_4)_2SO_4·FeSO_4·6H_2O 4. Fe_2(S0_4)_3

1. (b) 2. (c) +2 3. (b) dsp^2, square planar. 4. (d) It is diamagnetic with no unpaired electron. 5. (d) Fe_2(S0_4)_3 Explanation: (a) and (b) are coordination compounds hence cannot give free Fe ^ 2+ or Fe ^ 3+ ions in solution. (c) and (d) represent simple compounds hence are free to give ions in solution, but only Fe _2 ( SO _4 )_3 contains Fe ^ 3+ ions. ( NH _4 )_2 SO _4. FeSO _4. 6 H _2 O contains Fe ^ 2+ ions not Fe ^ 3+ ions.

Read the passage given below and answer the following questions: …

Read the passage given below and answer the following questions:
Consider the given sequence of reactions:

The following questions are multiple choice questions. Choose the most appropriate answer:

Identify W.

Compound Y is:

When X reacts with $CH_3COCl$ in presence of anhy. $AlCl_3$, the reaction is known as:

Fittig reaction.
Ullmann reaction.
Wurtz-Fittig reaction.
Friedel-Crafts acylation reaction.

When X is treated Ni-Al/ NaOH the product obtained is:

Benzene.
Phenol.
P-chlorophenol.
Triphenyl.

Compound Z is:

Phenol.
P-chlorophenol.
P-nitrophenol.
Nitrobenzene.

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Read the passage given below and answer the following questions:
Arrangement of ligands in order of their ability to cause splitting $(\Delta)$ is called spectrochemical series. Ligands which cause large splitting (large $\Delta$) are called strong field ligands while those which cause small splitting (small $\Delta$) are called weak field ligands. When strong field ligands approach metal atom/ ion, the value of $\Delta_0$ is large, so that electrons are forced to get paired up in lower energy $t_{2g}$ orbitals. Hence, a low-spin complex is resulted from strong field ligand. When weak field ligands approach metal atom/ ion, the value of $\Delta_0$ is small, so that electrons enter high energy $e_g$ orbitals rather than pairing in low energy $t_{2g}$ orbitals. Hence, a high-spin complex is resulted from weak field ligands. Strong field ligands have tendency to form inner orbital complexes by forcing the electrons to pair up. Whereas weak field ligands have tendency to form outer orbital complex because inner electrons generally do not pair up.
In these questions (Q. No. i-iv), a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

Assertion and reason both are correct statements and reason is correct explanation for assertion.
Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Assertion is correct statement but reason is wrong statement.
Assertion is wrong statement but reason is correct statement.

The following questions are multiple choice questions. Choose the most appropriate answer:

Assertion: In tetrahedral coordination entity formation, the d orbital splitting is inverted and is smaller as compared to the octahedral field splitting.

Reason: Spectrochemical series is based on the absorption of light by complexes with different ligands.

Assertion: In high spin situation, configuration of $d^5$ ions will be $\text{t}^3_{2\text{g}}\text{e}^2_\text{g}.$

Reason: In high spin situation, pairing energy is less than crystal field energy.

Assertion: $F^-$ ion is a weak field ligand and fonns outer orbital complex.

Reason: $F^-$ ion cannot force the electrons of $d_{z^2}$ and $d_{x^2-y^2}$ orbitals of the inner shell to occupy $d_{xy}, d_{yz}$ and $d_{zx}$ orbitals of the same shell.

Assertion: The crystal field model is successful in explaining the formation, structures, colour and magnetic properties of coordination compounds.

Reason: In spectrochemical series, ligands are arranged in a series of increasing field strength.

Assertion: $NF_3$ is a weaker ligand than $N(CH_3)_3$.

Reason: $NF_3$ ionizes to give $F^-$ ions in aqueous solution.

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Read the passage given below and answer the following questions: Carboxylic acids having an $\alpha$-hydrogen atom when treated with chlorine or bromine in the presence of small amount of red phosphorus gives $\alpha$-halocarboxytic acids. The reaction is known as Hell-Volhard-Zelinsky reaction. $\text{R}-\text{CH}_2-\text{COOH}+\text{X}_2\xrightarrow{\text{red p}}\text{R}-\text{CH}-\text{COOH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{X}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{X = Cl, Br)}$ When sodium salt of carboxylic acid is heated with soda lime it loses carbon dioxide and gives hydrocarbon with less number of C-atoms. $\text{R}-\text{COOH}\xrightarrow{\text{NaOH}}\text{R}-\text{COONa}\xrightarrow[\Delta]{\text{NaOH}+\text{CaO}}\text{R}-\text{H}+\text{Na}_2\text{CO}_3\\\text{Carboxylic}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Sod.}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Alkane}\\\ \ \ \ \ \text{acid}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{carboxylate}$ In these questions (Q. No. l-iv), a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

Assertion and reason both are correct statements and reason is correct explanation for assertion.
Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Assertion is correct statement but reason is wrong statement.
Assertion is wrong statement but reason is correct statement.

Assertion: $(CH_3)_3CCOOH$ does not give H.V.Z reaction.

Reason: $(CH_3)_3CCOOH$ does not have $\alpha$-hydrogen atom.

Assertion: H.V.Z. reaction involves the treatment of carboxylic acids having $\alpha$-hydrogens with $Cl_2$ or $Br_2$ in presence of small amount of red phosphorus.

Reason: Phosphorus reacts with halogens to form phosphorus trihalides.

Assertion: Propionic acid with $\frac{\text{Br}_2}{\text{P}}$ yields $CH_2Br - CHBr - COOH$.

Reason: Propionic acid has two $\alpha$-hydrogen atoms.

Assertion: $C_6H_5COCH_2COOH$ undergoes decarboxylation easily than $C_6H_5COCH_2COOH$.

Reason: $C_6H_5COCH_2COOH$ is $\beta$-keto acid.

Assertion: On heating 3-methylbutanoic acid with soda lime, isobutane is obtained.

Reason: Soda lime is a mixture of NaOH + CaO in the ratio 3 : 1.

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Read the passage given below and answer the following questions:
Metal carbonyl is an example of coordination compounds in which carbon monoxide (CO) acts as ligand. These are also called homoleptic carbonyls. These compounds contain both $\sigma$ and $\pi$ character. Some carbonyls have metal-metal bonds. The reactivity of metal carbonyls is due to (i) the metal centre and (ii) the CO ligands. CO is capable of accepting an appreciable amount of electron density from the metal atom into their empty $\pi$ or $\pi-\text{orbital}.$ These types of ligands are called $\pi-\text{accepter}$ or $\pi-\text{acid}$ ligands. These interactions increases the $\Delta_0$ value.
The following questions are multiple choice questions. Choose the most appropriate answer:

What is the oxidation state of metal in $[Mn_2(CO)_{10}]?$

$+1$
$-1$
$+2$
$0$

Among the following metal carbonyls, the $C-O$ bond order is lowest in:

$[Mn(CO)_6]^+$
$[Fe(CO)_5]$
$[Cr(CO)_6]$
$[V(CO)_6]^-$

Which of the following can be reduced easily?

$V(CO)_6$
$Mo(CO)_6$
$[Co(CO)_4]^-$
$Fe(CO)_5$

The oxidation state of cobalt in $K[Co(CO)_4]$ is:

$+1$
$+3$
$-1$
$0$

Structure of decacarbonyl manganese is:

Trigonal bipyramidial
Octahedral
Tetrahedral
Square pyramidal

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Read the passage given below and answer the following questions:
Amines are basic in nature. The basic strength of amines can be expressed by their dissociation constant, $K_b$ or $pK_b$.
$\text{RNH}_2+\text{H}_2\text{O}\rightleftharpoons\text{RNH}^+_3+\text{OH}^-$
$\text{k}_\text{b}=\frac{[\text{RNH}^+_3][\text{OH}^-]}{[\text{RNH}_2]}\text{and}\text{ pk}_\text{b}=-\log\text{k}_\text{b}$
Greater the $K_b$ value or smaller the $pK_b$ value, more is the basic strength of a mine. Aryl amines such as aniline are less basic than aliphatic amines due to the involvement of lone pair of electrons on N-atom with the resonance in benzene. In derivatives of aniline, the electron releasing groups increase the basic strength while electron withdrawing groups decrease the basic strength. The base weakening effect of electron withdrawing group and base strengthening effect of electron releasing group is more marked at p-position than at m-position. a-Substituted aniline is less basic than aniline due to ortho effect and is probable due to combination of electronic and steric effect.
The following questions are multiple choice questions. Choose the most appropriate answer:

Which of the following has lowest $pK_b$ value?

The strongest base among the following is:

$C_6H_5NH_2$
$p-NO_2 - C_6H_4NH_2$
$m-NO_2 - C_6H_4NH_2$
$C_6H_5CH_2NH_2$

Maximum $pK_b$ value of:

$(CH_3CH_2)_2NH$
$(CH_3)_2NH$

The order of basic strength among the following amines in benzene solution is:

Methylamine is more basic than $NH_3$.
Amines form hydrogen bonds.
Ethylamine has higher boiling point than propane.
Dimethylamine is less basic than methylamine.

$CH_3CH_2NH_2$ contains a basic $-NH_2$ group, but $CH_3CONH_2$ does not because:

Acetamide is amphoteric in character.
In ethylamine the electron pair on N-atom is delocalised by resonance.
In ethylamine there is no resonance while in acetamide the lone pair of electrons on N-atom is delocalised and is less available for protonation.
None of these.

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Read the passage given below and answer the following questions:
When a protein in its native form, is subjected to physical changes like change in temperature or chemical changes like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein.
The denaturation causes change in secondary and tertiary structures but primary structures remains intact. Examples of denaturation of protein are coagulation of egg white on boiling, curdling of milk, formation of cheese when an acid is added to milk.
The following questions are multiple choice questions. Choose the most appropriate answer:

Mark the wrong statement about denaturation of proteins.

The primary structure of the protein does not change.
Globular proteins are converted into fibrous proteins.
Fibrous proteins are converted into globular proteins.
The biological activity of the protein is destroyed.

Which structure(s) of proteins remains(s) intact during denaturation process?

Both secondary and tertiary structures.
Primary structure only.
Secondary structure only.
Tertiary structure only.

$\alpha$-helix and $\beta$-pleated structures of proteins are classified as:

Primary structure.
Secondary structure.
Tertiary structure.
Quaternary structure.

Cheese is a:

Globular protein.
Conjugated protein.
Denatured protein.
Derived protein.

Secondary structure of protein refers to:

Mainly denatured proteins and structure of prosthetic groups.
Three-dimensional structure, especially the bond between amino acid residues that are distant from each other in the polypeptide chain.
Linear sequence of amino acid residues in the polypeptide chain.
Regular folding patterns of continuous portions of the polypeptide chain.

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Read the passage given below and answer the following questions:
Although chlorobenzene is inert to nucleophilic substitution, however it gives quantitative yield of phenol when heated with aq. $Na OH$ at high temperature and under high pressure. As far as electrophilic substitution in phenol is concemed the — OH group is an activating group, hence, its presence enhances the electrophilic substitution at o - and p - positions.
The following questions are multiple choice questions. Choose the most appropriate answer:

Conversion of chlorobenzene into phenol involves:

Modified $S_N1$ mechanism.
Modified $S_N2$ mechanism.
Both (a) and (b).
Elimination-addition mechanism.

Phenol undergoes electrophilic substitution more readily than benzene because:

The intermediate carbocation is a resonance hybrid of more resonating structures than that from benzene.
The intermediate is more stable as it has positive charge on oxygen, which can be better accommodated than on carbon.
In one of the canonical structures, every atom (except hydrogen) has complete octet.
The — OH group is o, p-directing which like all other o, p - directing group, is activating.

Phenol on treatment with excess of cone. $HNO_3$ gives:

O - nitrophenol.
P - nitrophenol.
O - and p - nitrophenol.
2, 4, 6 - trinitrophenol.

Phenol is heated with a solution of mixture of $KBr$ and $KBrO_3$. The major product obtained in the above reaction is:

2 - bromophenol.
3 - bromophenol.
4 - bromophenol.
2, 4, 6 - tribromophenol.

The major product of the following reaction is:

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Read the passage given below and answer the following questions:
Aldehydes and ketones are reduced to primary and secondary alcohols respectively by $NaBH_4$ or $LiAlH_4$ as well as catalytic hydrogenation. The carbonyl group of aldehydes and ketones is reduced to

group on treatment with Zn-Hg and cone. HCl (Clemmensen reduction) or with hydrazine followed by NaOH or KOH in highly boiling solvent such as ethylene glycol (Wolff-Kishner reduction).Aldehydes differ from ketones in their oxidation reactions. Aldehydes are easily oxidised to carboxylic acids on treatment with $HNO_3, KMnO_4, K_2Cr_2O_7$ etc. Even mild oxidising agents mainlyTollens' reagent and Fehling's solution also oxidise aldehydes. Ketones are generally oxidised under vigorous conditions i.e., strong oxidising agents and at elevated temperatures, to give mixture of carboxylic acids having lesser number of C-atoms than the parent ketone.
The following questions are multiple choice questions. Choose the most appropriate answer:

Which of the following cannot be made by reduction of ketone or aldehyde with $NaBH_4$ in methanol?

1-Butanol
2-Butanol
2-Methyl-1-propanol
2-Methyl-2-propanol

The carbonyl compound producing an optically active product by reaction with $LiAlH_4$ is:

Propanone
Butanone
3-pentanone
Benzophenone

A substance $C_4H_{10}O (X)$ yields on oxidation a compound $C_4H_8O$ which gives an oxime and a positive iodoform test. The substance X on treatment with cone. $H_2SO_4$ gives $C_4H_8$. The structure of the compound (X) is:

$CH_3CH_2CH_2CH_2OH$
$CH_3CH(OH)CH_2CH_3$
$(CH_3)_3COH$
$CH_3CH_2- O - CH_2CH_3$

In the oxidation of by acidified $K_2Cr_2O_7$, the products are:

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\ ^\text{14}\text{C}-\text{OH}$ and $\text{CH}_3\text{CH}_2\text{COOH}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3(\text{CH}_2)_2\text{COOH}-\text{C}-\text{OH}$ and $ \ \ \ \ \ \ \ 14\\\text{CH}_3\text{CH}_2\text{COOH}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 14\\\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}+\text{HCOOH}$
None of these.

The appropriate reagent for the following transformation is:

$\text{Na}_2\text{NH}_2,^-\text{OH}$
$\text{NaBH}_4$
$\frac{\text{H}_2}{\text{Ni}}$
$\text{AICl}_3$

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Read the passage given below and answer the following questions:
In haloalkanes, when a nucleophile stronger than the halide ion approaches the positively charged carbon atom of an alkyl halide, the halogen atom along with its bonding electron pair gets displaced and a new bond with the carbon and the nucleophile is formed. These reactions are called nucleophilic substitution reactions.

In these reactions the atom or group of atoms which loses its bond from carbon and takes on an additional pair of electrons is called leaving group. Halide ions are good leaving groups. Some important nucleophilic substitution reactions ofhaloalkanes with common nucleophiles are given in the table below.

	Reagent	Nucleophile (Nu^-)	Substitution product R-Nu	Class of main product
1.	$NaOH$ or $KOH$ or moist $Ag_2O$	$^-OH$	$ROH$	Alcohol
2.	$H_2O$	$H^2O$	$ROH$	Alcohol
3.	$Nal$	$I^-$	$R – I$	Alkyl iodide
4.	$R'NH_2$	$\text{R'}\ddot{\text{N}}\text{H}_2$	$RNHR'$	Sec. amine
5.	$KCN$	$\overline{\text{C}}\equiv\text{N}:$	$RCN$	Nitrile (cyanide)
6.	$KNO_2$	$O = N – O^-$	$R – O – N = O$	Alkvl nitrite

In these questions (Q. No. i-iv), a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

Assertion and reason both are correct statements and reason is correct explanation for assertion.
Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Assertion is correct statement but reason is wrong statement.
Assertion is wrong statement but reason is correct statement.

Assertion: Alkyl halides are hydrolysed to alcohols by moist silver oxide.

Reason: $RCI$ is hydrolysed to $ROH$ easily but reactions slow down on addition of KI.

Assertion: Alkyl halides fonn alkenes when heated above $300^\circ C.$

Reason: $CH3CH21$ reacts slowly with strong base as compared to $CD_3CH_2I.$

Assertion: RBr reacts with $AgNO_2$ to give nitroalkane.

Reason: Silver nitrite $(AgNO_2)$ is an ionic compound, therefore the negative charge on nitrogen is the attacking site.

Assertion: The nucleophilic substitution of vinyl chloride is difficult than ethyl chloride.

Reason: Vinyl group is electron donating group.

Assertion: $2-$Bromobutane on reaction with sodium ethoxide in ethanol gives 1-butene as the major product.

Reason: $1-$Butene is less stable than $2-$butene.

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Read the passage given below and answer the following questions:
All the elements of group $16$ have $ns^2np^4$ configuration in their outermost shell. Therefore, the atoms of these elements try to gain or share two electrons to achieve noble gas configuration. Sulphur and other elements of group $16$ are less electronegative than oxygen, so, they cannot accept electrons easily. By sharing of two electrons with other elements, these elements acquire $ns^2np^6$ configuration and exhibit $+2$ oxidation state. Except oxygen, group $16$ elements have vacant d-orbitals in their valence shell to which electrons Call be promoted from P- and s-orbitals of the same shell As a result, they Call show $+4$ and $+6$ oxidation states also.
The following questions are multiple choice questions. Choose the most appropriate answer:

Oxygen shows $+2$ oxidation state in.

$OF_2$
$H_2O$
$CI_2O$
$H_2O_2$

Like sulphur, oxygen is not able to show $+4$ and $+6$ oxidation states because?

Oxygen is a gas while sulphur is a solid.
Sulphur has high ionisation enthalpy as compared to oxygen.
Oxygen has nod-orbitals in its valence shell.
Oxygen has high electron affmity as compared to sulphur.

Compounds of sulphur with $+4$ oxidation state acts as a/ an.

Oxidising agent.
Reducing agent.
Both oxidising as well as reducing agen.
Cannot be predicted.

Oxidation state of sulphur in $Na_2S_4O_6$ is:

$\frac{7}{2}$
$\frac{5}{2}$
$\frac{1}{2}$
$\frac{3}{2}$

The oxidation states of sulphur in $S_8, SO_3$ and $H_2S$ are respectively.

$0, +6$ and $-2$
$+6, 0$ and $-2$
$-2, 0$ and $+6$
$+2, +6$ and $-2$

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