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malfutka [58]
3 years ago
11

Of the substances xe, ch3cl, hf, which has: (a) the smallest dipole–dipole forces? (b) the largest hydrogen bond forces? (c) the

largest dispersion forces?
Chemistry
2 answers:
frosja888 [35]3 years ago
6 0
A) <span>he smallest dipole–dipole forces have xenon (Xe), because xenon is a noble gas and have stable electron configuration and do not form dipole.
b) t</span><span>he largest hydrogen bond forces have HF because hydrogen bonds are formed between hydrogen and F, O and N.
</span>c) the largest dispersion forces have CH₃Cl because he has strongest <span>induced temporary dipoles.</span>
Pavel [41]3 years ago
6 0

Answer:

(a) Xe

(b) HF

(c) Xe

Explanation:

<u>Dipole-dipole interaction:- </u>

This is the interaction between the two dipoles in the molecule which is formed by varying electronegativities.

<u>Hydrogen bonding:- </u>

Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.

Partially positive end of the hydrogen atom is attracted to partially negative end of these atoms which is present in another molecule. It is strong force of attraction between the molecules.

<u>London dispersion forces:- </u>

The intermolecular force acting in the molecule are induced dipole-dipole forces or London Dispersion forces / van der Waals forces which are the weakest intermolecular force.

<u>Thus, Xe will have the smallest dipole–dipole forces because there is no chances of forming a dipole. Also, it has the largest dispersion forces as the other both has dipole and involved in other interactions.</u>

<u>Also, H-F bond is more polar than C-Cl bond and thus HF has the largest hydrogen bond forces.</u>

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pH is the hydrogen ion concentration and pOH is the hydroxide ion concentration in the solution. pH KOH is 12.73, pOH KOH is 2.24 and pH NaCl is 7.

<h3>What are pH and pOH?</h3>

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The relation between the pH and pOH can be given as, \rm pOH = 14 - pH

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\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 87 \times  10^{-13}\;\rm  M ]\\\\&= 12.73\end{aligned}

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pOH of KOH can be calculated by the formula,

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Substituting value in the equation:

\begin{aligned} \rm pOH &= \rm -log [OH^{-}]\\\\&= \rm -log [5. 81 \times 10^{-3}\;\rm  M ]\\\\&= 2.24 \end{aligned}

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The pH of NaCl can be calculated by the formula,

\rm pH = \rm -log [H^{+}]

In the third case, the concentration of the NaCl is 1. 00\times 10^{-7}\;\rm  M

Substituting values in the equation:

\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 00 \times  10^{-7}\;\rm  M ]\\\\&= 7 \end{aligned}

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