Chemistry Test - 30

Platinum generally occurs in free state.

Platinum is a chemical element with the symbol Pt and atomic number 78. It is a dense, malleable, ductile, highly unreactive, precious, silverish-white transition metal. Since it is noble metal it is found in free state.

The law of octaves was found to be applicable till calcium.

After calcium, the properties of the next elements were not synchronous as per the law of octaves. Law of octaves, in chemistry, the generalization made by the English chemist J.A.R. Newlands in 1865 that, if the chemical elements are arranged according to increasing atomic weight, those with similar physical and chemical properties occur after each interval of seven elements.

The \(\mathrm{C}-\mathrm{Cl}\) bond in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}\) is shorter than in \(\mathrm{CH}_{3} \mathrm{Cl}\) and therefore, stronger.

• The carbon in \(\mathrm{C-Cl}\) bond in chlorobenzene is \(\mathrm{sp}^{2}\) hybridised, while in \(\mathrm{CH}_ 3-\mathrm{Cl}\) is \(\mathrm{sp}^{3}\) hybridised.
• In \(\mathrm{sp}^{2}\) hybrid orbitals have more of \(\mathrm{s}\)-character and hence the carbon if chlorobenzene withdraws the electron pair between \(\mathrm{C-Cl}\) with greater force.
• As a result, \(\mathrm{C-Cl}\) bond is shorter than \(\mathrm{CH}_{3}-\mathrm{Cl}\).
• The lone pair on chlorine are dispersed throughout the benzene ring by resonance.
• This gives the \(\mathrm{C-Cl}\) bond a double bond character.

At

(a) \(\rightarrow \mathrm{CO}_2\) exist as gas

(b) \(\rightarrow\) liquefaction of \(\mathrm{CO}_2\) starts

(c) \(\rightarrow\) liquefaction ends

(d) \(\rightarrow \mathrm{CO}_2\) exist as liquid.

Between (b) \& (c) \(\rightarrow\) liquid and gaseous \(\mathrm{CO}_2\) co-exist.

As volume changes from (b) to (c) gas decreases and liquid increases.

(A), (C) \(\rightarrow\) correct

\(\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \)

\(\Rightarrow \mathrm{mv}=\frac{\mathrm{h}}{\lambda}=\frac{6.626 \times 10^{-34} \mathrm{~kg} \frac{\mathrm{m}^2}{\mathrm{sec}^2} \times \mathrm{sec}}{3.3 \times 10^{-10} \mathrm{~m}} \)

\(\mathrm{mv}=\frac{6.626 \times 10^{-24}}{3.3}=2 \times 10^{-24} \mathrm{~kg} \mathrm{~m} \mathrm{sec}-1 \)

\(\text { Kinetic energy }=\frac{1}{2} \mathrm{mv}^2 \)

\(=\frac{(\mathrm{mv})^2}{2 \mathrm{~m}} \)

\( =\frac{\left(2 \times 10^{-24}\right)^2}{2 \times 9.1 \times 10^{-31} \mathrm{~kg}} \)

\(=2.18 \times 10^{-18} \mathrm{~J} \)

\(=21.8 \times 10^{-19} \mathrm{~J}\)

Total energy absorbed = lonization energy + Kinetic energy

\(=(21.76+21.8) \times 10^{-19} \)

\(=43.56 \times 10^{-19} \mathrm{~J} \)

\(\approx 2 \text { times of } 21.76 \times 10^{-19} \mathrm{~J}\)

Elements of group \(13\) are part of p-block elements.

The last electron enters in the outermost p-orbital in the p block elements, from group \(13\) to group \(18\) the entire elements belong to p-block, whereas group \(13\) is called a boron family. It includes the elements boron, aluminum, gallium, indium and thallium.

Given:

\(a=390~pm\)

For bcc structure:

Interionic distance \(=r^{+}+r^{-}=\frac{\sqrt{3}}{2} a\)

where, \(a:\) edge length

\(r_{N H_{4}^{+}}^{+}+r_{Cl^{-}}^{-}=\frac{\sqrt{3}}{2} a\)

\(r_{N H_{4}^{+}}^{+}+180=\frac{\sqrt{3}}{2} \times 390\)

\(r_{N H_{4}^{+}}^{+}+180 = 338\)

\(r_{N H_{4}^{+}}^{+}=338 {~pm}-180 {~pm}\)

\(r_{N H_{4}^{+}}^{+}=158 {~pm}\)

A thermodynamic state function is a quantity whose value is independent ofthe path.Its value depends on initial and final states and is independent of the path followed. Functions like \(p, V, T\), etc. depend only on the state of a system and not onthe path.A state function describes the equilibrium state of a system.

The atomic number is equal tonumber of protons \(= 16\). The element issulphur (S).

Atomic mass number \(=\) number ofprotons \(+\) number of neutrons

\(= 16 + 16 = 32\)

Species is not neutral as the number ofprotons is not equal to electrons. It isanion (negatively charged) with chargeequal to excess electrons \(= 18 – 16 = 2\)

Thus symbol is \({ }_{16}^{32} \mathrm{~S}^{2-}\).

The co-ordination number and oxidation state of \(\mathrm{Cr}\) in \(\mathrm{K}_{3}\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]\) are 6 and +3 respectively.

Co-ordination number of \(\mathrm{Cr}^{3+}\) ion is six because chromium ion is surrounded by three bidentate ligands.

For oxidation state of central metal ion \(3+x-6=0\)

[where \(x\) is the oxidation state of central metal ion]

As oxalate is a bidentate ligand, coordination number of Cr is 6.

Since the overall charge on the complex is 0, the sum of oxidation states of all elements in it should be equal to 0

Let the oxidation number of Cr be x.

Therefore, 3+x+3×(−2)=0 \(\Rightarrow\)x=+3