Halogens (Group 17) - Properties and Trends

Hydrolysis of $NCl$ gives $HOCl$ (hypochlorous acid)

$TeF _6$ canbe easily hydrolysed because of large size of Te central atom and low electronegativity the bond polarity of this bond is more .so, it is more reactive than other compounds like $SF _6$

PbI4 is the least stable because Pb(IV) is a strong oxidizing agent and I- is a strong reducing agent, leading to spontaneous reduction to Pb(II) and I2. In contrast, fluorides are more stable because fluorine is a poor reducing agent.

PbI4 does not exist because the Pb(IV) state is unstable relative to Pb(II) in the presence of a strong reducing agent like iodide. The energy released by the reduction of Pb(IV) to Pb(II) is greater than the energy required to form the Pb-I bonds.

Halogens have high ionisation energies, high electronegativities and high electron affilities due to their smallest size in a particular period. As size decreases ionisation energy, elctronegativity and electron affinity increaes

$KI+I _2\rightarrow KI _3$ soluble 
$\therefore $ Iodine is more soluble in KI (aq)

$KI\ (aq)\rightarrow K^{(+)}+I^-$

The reaction of $I _2$ with water is endothermic 
so, change in free energy is Positive.

Astatine has many isotopes, all of which are radioactive in nature. Their half-lives are also very small. Hence, astatine has no stable isotopes.

Van der Waals force- It is a general term used to define the attraction of intermolecular forces between molecules.

As moving down the group electron affinity of elements decreases

But in case of $\textbf{Fluorine}$, it is a small atom with a small amount of space available in its 2p orbital. Because of this, any new electron trying to attach to fluorine experiences lower electron affinity from the electrons already living in the element's 2p orbital. Since chlorine's outermost orbital is a 3p orbital, there is more space, and the electrons in this orbital are inclined to share this space with an extra electron. Therefore, chlorine has a higher electron affinity than fluorine, and this orbital structure causes it to have the highest electron affinity of all of the elements.

Hence electron affinity of $\textbf{Chlorine}$ is higher in halogens.

Chlorine displaces Bromines and Iodine from their salts as chlorine is more electronegative than bromine and iodine.
$2KBr+Cl _2\rightarrow 2KCl+Br _2$
$2KI+Cl _2\rightarrow 2KCl+I _2$

Reaction of $F _2$ and $H _2O$-

Iodine crystals sublime because though there are covalently bonded within the molecule holding two iodine atoms together, only Van der Waals' forces exist between the iodine molecules. Van der Waal's forces are very weak interactions between molecules of a substance hence the Iodine molecules can escape from the solid structure easily.

Similar properties:-

The percentage of s-character increases with the electronegativity of the central atom or the nature of the hybrid orbitals. For NO2+, NO3-, and NH4+, the hybridization and bond angles dictate the s/p character distribution, and the listed order is incorrect.

Volatility depends on intermolecular interactions of a compound, as molecular weight increases so do the Vander wals forces. So the expected order should be: $HF < HCl<HBr<HI$ but we must remember HF has hydrogen bonding and the extent of Hydrogen bonding is very high because of high electonegativity of fluorine and linear bonding. Hence HF is least volatile. The correct order is $HCl>HBr>HI >HF$.

When concentrated $HBr$ is heated with $KMnO _4$ then bromine is liberated.The reaction involved is $2KMnO _4+16HBr\longrightarrow 2MnBr _2=2KBr+8H _2O+5Br _2$

Flourine is the most reactive element in the halogen group. 

Down the group, metallic nature increases. If metallic nature increases, basic nature increases. Among halogens, $I _2$ is at bottom and it has high metallic nature than other halogens. So, it is more basic.

HF is actually a weak acid compared to other hydrogen halides due to strong hydrogen bonding and high bond dissociation enthalpy, making the statement that HF is stronger than HCl incorrect.

According to Fajans rule larger the size of anion,more covalent will be the bond . So, according to this most covalent halide is iodide as because it is the least electronegative among all the halides,greater size of halide anion,so in a covalent compound it has the least polarising

$\text{The volatility of the Group 17 elements decrease down the group due to the increse the numnber of electrons.}$

The given statement is True. 
In Japan Sunami, there has been increase in radioactive iodine content in salt.
Traces of iodine are found in mineral deposits. Iodine also finds its way into the animal system as thyroxin in thyroid glands.

Tokyo, the city first reported a spike of radioactive iodine in its water supply on March 23, 2011 at levels below the limit for adults but above the threshold of safety for infants; since then, iodine levels in Tokyo's water have declined below both thresholds. 

Dioxygenyl hexafluoroplatinate $(O _2^+[PtF _6]^-)$ is a orange coloured salt. It is a solid and is used to prepare co-ordination complexes.

Catenation,  chemical linkage into chains of atoms of the same element, occurring only among the atoms of an element that has a valence of at least two and that forms relatively strong bonds with itself. The property is predominant among carbon atoms, significant among sulfur and silicon atoms, and slightly present among germanium, nitrogen, selenium, and tellurium atoms.
here, apart from iodine all other have this property.

Sublimation is the transition of a substance directly from the solid to the gas phase without passing through the intermediate liquid phase..
At room temperature, fluorine and chlorine are gases, bromine is liquid, and iodine is solid. They form diatomic molecules in the pure state. Since, $I _2$ is solid at room temperature. It will be sublimated and remaining can not be sublimated
Hence, options D is correct.

Bond dissociation energy decrease
down the group
But bonds dissociation energy of 
$F _2$ is less than excepted due to 
large repulsion of $e^-$ in $F _2$

There are two factors that determine the reactivity of the halogenation of alkanes.
1. Lesser the positive value of $E _{act}$ of the first chain propagating step, more reactive is the halogen.
2. Higher the negative value of overall heat of reaction $(\Delta H _{\zeta}^{})$ more reactive is the halogen. This explains high reactivity and explosive violence with which $F _2$ reacts with $CH _4$.

Iodine reacts with starch to produce a blue/black color.

Base is an electron pair donor. The more available the electrons, the more readily they can be donated to form a new bond to the proton and, and therefore the stronger base. When comparing atoms within the same group of the periodic table, the larger the atom the weaker the H-X bond and the lower the electron density making it a weaker base.

Iodine's main role in animal biology is as constituents of the thyroid hormones, thyroxine (T4) and triiodothyronine (T3). These are made from addition condensation products of the amino acid tyrosine, and are stored prior to release in an iodine-containing protein called thyroglobulin. T4 and T3 contain four and three atoms of iodine per molecule, respectively. The thyroid gland actively absorbs iodine from the blood to make and release these hormones into the blood, actions which are regulated by a second hormone TSH from the pituitary. Thyroid hormones are phylogenetically very old molecules which are synthesized by most multicellular organisms, and which even have some effect on unicellular organisms.

Molten $KHF _{2}$ is used in the preparation of fluorine.

Tincture of iodine contains 0.5 ounce iodine,  0.25 ounce KI and 1 pint of rectified spirit.

$Br^{-}$ can reduce $Cl _{2}$ to $Cl^{-}$ $(E _{cell}=+ve)$

$I^{-}$ can reduce $Br _{2}$ to $Br^{-}$ $(E _{cell}=+ve)$

$E _{cell}=+ve$ in both case

$\therefore$ Reaction is feasible

The iodine can be obtained by reacting chlorine or bromine by NaI. This is because the electronegativity of chlorine and bromine is more than the iodine and also the reactivity of chlorine and bromine is more than the iodine.

Reactivity order:-
$F _2>Cl _2> Br _2>I _2$
Thus, chlorine is more reactive than bromine and bromine is more reactive than iodine.

In KI solution, $I _2$ readily dissolves and forms $KI _3$ which contains polyhalide ion $I _3^-$

The reactivity of halogens decreases with increase in atomic number of halogens. So the correct order of reactivity of halogens is $F _2 > Cl _2 > Br _2 > I _2.$

Iodide of MilIon's base is $\displaystyle NH _2HgO.HgI$
It is formed when Nessler's reagent (alkaline solution of $\displaystyle K _2[HgI _4]$) reacts with $NH _3$. 

Iodised salt is table salt mixed with a minute amount of one of the four iodine containing salts of hydroiodic acid or iodic acid: potassium iodate or potassium iodine and sodium iodate or sodium iodide.

Iodine is less soluble in water than in carbon tetrachloride. Iodine is a non polar compound.
Iodine, being a non-polar compound is more soluble in $CCl _4$ (non-polar compound) than in water because "like dissolve like". So both the assertion and reason are false.

Elements of halogen family are in colored gaseous state, in liquid state and with metallic properties.
At room temperature, fluorine and chlorine are gases, bromine is a liquid and iodine is a solid.
The colours of fluorine, chlorine, bromine and iodine are yellow, greenish yellow, red and violet.
Iodine has metallic luster.

Halogens have similar reactivity because they have similar outer shell electron configurations of $\displaystyle  ns^2np^5$. 
Thus, all halogens gain one electron to complete octet. Thus, they form uni-negative ion (anion).

A purple solid (iodine) is produced by treatment of the dry solid $KI$ with a mild oxidizing agent. The mild oxidising agent oxidizes iodide ion to iodine.

Halogens have 7 electrons in their valence shell.

Element Z has electronic configuration ${1s}^2 2s^2 2p^{5}$ , its belongs to 17th group so, it can form compound $MgZ _2$.

Iodine can be displaced from potassium iodide if a more reactive element is made to react with potassium iodide. In the presence of a mild oxidising agent, the iodine ions are oxidised to $I _2$. Iodine is purple in colour.

The gas that would form when hydrobromic acid is combined with nickel (II) sulfide is $H _2S$

When $KClO _3$ is reacted with $MnO _2$ in presence of heat it forms $KCl,O _2,MnO _2$ respectively.

As the oxidation number of chlorine(non metal) increases in the given acids the acidic nature of the compound increases. So the oxidation numbers of chlorine in the given compounds are $+1,+3,5,+7$ respectively.

A one litre flask is full of brown bromine vapour. Intensity of brown colour vapour will not decrease a appreciable amount on adding to flask some of Piece of marble.

Hence option A is correct.

Affinity do not have any relation with the efficiency of oxidizing ability.

The bromine releases iodine from the potassium iodide Iodine reacts with starch to produce a blue color.

$Br _2+KI\rightarrow KBr+I _2$

Hence option A is correct

Option B and C reactions are not possible because the bromine do not displace chlorine and chlorine do not displace florine as chlorine and florine are most stable than bromine and chorine in their compounds respectively.

Reaction of iodine with fuming nitric acid is given by 

The bond dissociation energies of $I _2$, $Br _2$ and $Cl _2$ increase in the following order:

Potassium bromide ($KBr$) is a salt, widely used as an anticonvulsant and a sedative in the late 19th and early 20th centuries.

As we move down the group acidic strength decreases as more is the electronegativity of the central atom the shared pair of $e^-$ will be more towards the central atom; resulting in the release of Hydrogen easy.

Method is for enriching $Br _2$:

$2\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O} _{4}+2\mathrm{K}\mathrm{O}\mathrm{H}\rightarrow 2\mathrm{K} _{2}\mathrm{M}\mathrm{n}\mathrm{O} _{4}+\mathrm{H} _{2}\mathrm{O}+\mathrm{O}$
$\frac{2\mathrm{K} _{2}\mathrm{M}\mathrm{n}\mathrm{O} _{4}+2\mathrm{H} _{2}\mathrm{O}\rightarrow 2\mathrm{M}\mathrm{n}\mathrm{O} _{2}+4\mathrm{K}\mathrm{O}\mathrm{H}+2\mathrm{O}}{2\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O} _{4}+\mathrm{H} _{2}\mathrm{O}\xrightarrow[]{alkalme}2\mathrm{M}\mathrm{n}\mathrm{O} _{2}+2\mathrm{K}\mathrm{O}\mathrm{H}+3[\mathrm{O}]}$
$\frac{KI+[\mathrm{O}] \rightarrow \mathrm{K}\mathrm{I}\mathrm{O} _{3}}{2\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O} _{4}+\mathrm{K}\mathrm{I}+\mathrm{H} _{2}\mathrm{O}\rightarrow 2 KOH +2\mathrm{M}\mathrm{n}\mathrm{O} _{2}+\mathrm{K}\mathrm{I}\mathrm{O} _{3}}$
$KI+3[O]\rightarrow KIO _3^{-}$
Hence option A is correct.

Hypohalites $(XO^-)$ disproportionates in aqueous solution on heating to produce halide ion and halate ion.

Iodine
solutions are prepared by dissolving $I _2$ in a concentrated solution of potassium
iodide ($KI$).
Hence option D is correct.

The colour of the $\displaystyle X _2$ molecules of group 17 elements changes gradually from yellow to violet down the group. This is due to  (B) Decrease in $\displaystyle {\pi}^{\ast} - {\sigma}^{\ast}$ gap down the group and (C) decrease in HOMO−LUMO gap down the group. Halogens absorb visible light and appear coloured. Outer electrons are excited to higher energy level. Smaller halogen has higher excitation energy and vice versa.

The reduction potential of the substance is the ability of the substance to be reduced.So as the reduction potential increases reducing ability of the  substance increases.It means oxidizing power of the substance (The ability of the substance to make the other substance to lose the electrons increases which noting but oxidizing power) increases.

So the decreasing order of oxidizing power is $Br{ O } _{4}^{-}$ > $I{ O } _{4}^{-}$ > $Cl{ O } _{4}^{-}$.