Chemistry Test - 45

In halogens the order of electronegativity of their acid is \(F >Cl >Br >I\).

Among these acids, \(Cl\) is the most electronegative so its acid with \(O - H\) is more stable with a negative charge.

(I) HOCl (Strong acid), (II) HOBr, (III) HOI (weak acid)

So, the correct order of decreasing acid strength is \(I>II>III\)

(A) \(\mathrm{Ne}_2\left(20 \mathrm{e}^{-}\right)\)

\(\Rightarrow \sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \sigma 2 \mathrm{p}_{\mathrm{z}}{ }^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2 \pi 2 \mathrm{p}_{\mathrm{y}}{ }^2 \pi^* 2 \mathrm{p}_{\mathrm{x}}{ }^2 \pi^* 2 \mathrm{p}_{\mathrm{y}}{ }^2 \sigma^* 2 \mathrm{p}_{\mathrm{z}}{ }^2\)

Bond order \(=\frac{10-10}{2}=0 \Rightarrow\) (iii) of List-II.

(B) \(\mathrm{N}_2\left(14 \mathrm{e}^{-}\right)\)

\(\Rightarrow \sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \pi 2 \mathrm{p}_{\mathrm{x}}^2, \pi 2 \mathrm{p}_{\mathrm{y}}^2 \sigma 2 \mathrm{p}\) Bond order \(=\frac{10-4}{2}=3 \Rightarrow\) (iv) of List-II.

(c) \(\mathrm{F}_2\left(18 \mathrm{e}^{-}\right)\)

\(\Rightarrow \sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \sigma 2 \mathrm{p}_{\mathrm{z}}{ }^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2 \pi 2 \mathrm{p}_{\mathrm{y}}{ }^2 \pi^* 2 \mathrm{p}_{\mathrm{y}}{ }^2 \pi^* 2\)

Bond order \(=\frac{10-8}{2}=1 \Rightarrow(\) i) of List-II.

(d) \(\mathrm{O}_2\left(16 \mathrm{e}^{-}\right)\)

\(\Rightarrow \sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \sigma 2 \mathrm{p}_{\mathrm{z}}{ }^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2 \pi 2 \mathrm{p}_{\mathrm{y}}{ }^2 \pi^* 2 \mathrm{p}_{\mathrm{x}}{ }^1 \pi^* 2 \mathrm{p}_{\mathrm{y}}{ }^1\)

Bond order \(=\frac{10-6}{2}=2 \Rightarrow\) (ii) of List-II.

The correct matching is option (a).

No. of electrons in antibonding

MO (Molecular Orbital)

Note : Bond order \(=\frac{- \text { No. of electrons in bonding MO }}{2}\)

A catalyst works by providing an alternative reaction pathway to the reaction product. The rate of the reaction is increased as this alternative route has lower activation energy than the reaction route not mediated by the catalyst.

In \(A g_{2} S O_{4}\), silver is present as \(A g^{+}\), in \(C u F_{2}\), copper is present as \(C u^{2+}\), in \(Z n F_{2}\), zinc is present as \(Z n^{2+}\), in \(C u_{2} C l_{2}\), the copper is present as \(C u^{+1}\).

As they are metal cations, the electrons are removed from their actual configuration to form ions.

The respective transition elements exists in:

\(A g^{+}-[A r]4 d^{10} 5 s^{0},\) (have completely filled \(d\) orbital).

\(C u^{+2}-[A r]3 d^{9} 4 s^{0},\) (have incompletely filled \(d\) orbital).

\({Zn}^{+2}-[A r]3 d^{10} 4 s^{0},\) (have completely filled \(d\) orbital).

\({Cu}^{+1}-[A r]3 d^{10} 4 s^{9},\) (have completely filled \({d}\) orbital).

Since, \(C u^{+2}\) has unpaired electron, it will show electron transitions. Therefore, will be coloured.

A well-stoppered thermos flask contains some ice cubes. This is an example of an isolated system.

An isolated system is a thermodynamic system that can not exchange either energy or matter outside the boundaries of the system.In a thermos, ice can be kept for many hours because the walls of the thermos are insulated and it does not allow heat to flow out of the flask.

The functional group present in organic, acid is –COOH.

The organic acids are known as carboxylic acid as the functional group is carboxyl group. This is the structure of the organic acid where R can be an alkyl or benzyl group and –COOH is the functional group. They are known as carboxylic acids. Their general formula is C_nH_2nO₂.

Transition metal compounds are usually coloured. This is due to the electronic transition within the d-orbitals.

Transition elements (also known as transition metals) are elements that have partially filed d orbitals.

Many ionic and covalent compounds of transition elements are coloured and this colouring property is due to \(d-d\) electronic excitation. The colour of the compound is complementary to the colour absorbed.

Maleic acid is stronger than fumaric acid.

• Maleic acid is cis-butenedioic acid whereas fumaric acid is trans-butenedioic acid.
• Maleic acid is able to lose \(H^{+}\)ion and it results in the formation of intra hydrogen bond. Whereas fumaric acid being a trans isomer has strong interaction with both the oxygen atoms of each carboxylic group. Therefore, fumaric acid is unable to give hydrogen ions as compared to maleic acid.
• That is why, maleic acid is stronger than fumaric acid.

BF₃-

• This moleculedoes nothave its octetfulfilled.
• To fill its octet, it canaccepta lone pair and act as aLewis acid.

NH₃-

• Nitrogen has anextra lone pairon its head.
• To get rid of it it candonatethis pair acting as aLewis base.

The interaction-

• Lewis acid has avacant unoccupied orbital called LUMO.
• Lewis bases have fulfilled occupiedorbital called HOMO.
• Electron density from HOMO is donated toLUMOof Lewis acid.
• A simple HOMO and LUMOi nteraction is shown below:

• Thisdonationof electron density forms acovalent coordinate betweenNH₃and BF₃.

Aco-ordinate bond exists between NH₃ and BF₃.

We know a negatively charged gold solution is coagulated by a Magnesium Chloride solution readily than positively charged iron (III) hydroxide solution. Also, in Magnesium chloride, the charge on cation is twice the charge on the anion. Higher is the charge on ion, higher will be its coagulating power.

Also we have studied that Sodium sulphate solution causes coagulation in both solutions. Negatively charged gold sol is coagulated by sodium ions and positively charged iron (III) hydroxide sol is coagulated by sulphate ions.

The dispersed phase in colloidal gold and colloidal iron (III) hydroxide is negatively and positively charged respectively. Mutual Coagulation is affected when two solutions are mixed.. This is due to neutralization of the charge on one sol by opposite charge on another sol.

By electrophoresis, coagulation in both the solutions can be brought. We know that, during electrophoresis, through the colloidal solution an electric current is passed and the experiment is arranged in such a manner that only particles can move.

After studying above points, we can conclude that option A, B and D are correct and C is incorrect.