Explanation
H2O because of hydrogen bonding is liquid and has highest boiling point.
Molecular shapes of SF4 ( See –saw ), CF4 ( tertrahedral ) and XeF4 (square planer) are different with 1, 0 and 2 lone pair of electrons on S, C and Xe respectively .
Boiling point of H2O is the highest and that of H2S is the lowest. In case of H2Te because of larger size as compared to H2Se , the vander Waals attraction is stronger and thus the b.p is higher than H2Se.
Noble gases constitute about 1% of the total volume of air. The remaining 99% volume in mainly due to oxygen and nitrogen. When 1 litre (1000 ml ) of air is passed over heated copper,oxygen is consumed and on further passing over magnesium, nitrogen in consumed according to the equations:
2Cu + O2 → 2CuO
3Mg + N2 → Mg3N2
In this way about 990 ml of air is consumed. The remaining volume of about 10ml is due to mainly Ar and other noble gases
Nitrogen gas prepared from air contains Ar as an impurity which is present in the atmosphere most abundantly
The atomic radius fluorine is less than that of the next element, noble gas (neon) in which case only van der Waals radius is considered which is larger than covalent radius
The most abundant rare gas found in the atmosphere is argon and not helium.
Xe in XeO3 is sp3- hybridized having one lone pair of electrons on Xe atom giving a trigonal pyramidal structure
XeF4 has a square planer structure, XeF6 has distorted octahedral, XeO3 has a trigonal pyramidal and XeO4 has a tetrahedral structure.
In XeF4. Xe is sp3 d2- hybridized with two lone pairs of electrons giving a square planar structure.
The number of lone pairs of electrons on Xe atom in XeF2, XeF4 and XeF6 are 3, 2 and 1 respectively
He II is liquid form of helium which is obtained on cooling gaseous He to 2.2 K at one atomosphere pressure.It is able to flow uphill like a gas and thus has properties of a degenerate gas.
KrF2 is a F‒ donor and forms complexes with F‒ acceptors where only cationic species of Kr will be present.
(d) Cl2O7
Conc. H2SO4 oxidises HI to I2 and therefore, the reaction of KI with conc. H2SO4 can not give HI. KI + H2SO4 → KHSO4 + HI
H2SO4 + 2HI → SO2 + I2 + 2H2O
Because of the absence of d-orbitals and highest electronegativity, F does not show positive oxidation state.
F (At. No. 9) is the most electronegative element. Its electronic configuration is 1s22s22p5.
Bond dissociation energy increases in the order as F2 < Cl2 < O2 < N2 158 243 498 946 (kJmol-1)
Cl2 gas reacts with dry slaked lime, Ca (OH)2 to give bleaching powder
Ca(OH)2 + Cl2 → CaOCl2 + H2O
Conc. HNO3 reacts with I2 to give HIO3 or HOIO2
I2 + 10HNO3 → 2HIO3 + 10NO2 + 4H2O
Due to non-availability of d-orbitals F does not show positive oxidation states.
All these are isoelectronic species. Their sizes increase as the nuclear charge decreases from Mg to N.
As halogens have seven electrons (ns2np5) in the valence shell, they have a strong tendency to acquired the nearest inert gas configuration by gaining an electron from the metallic atom and form halide ions easily.
Compound OF2 OH2 OCl2 ClO2 Bond angle 103° 105° 111° 118° In all these compounds, the central atom is sp3-hybridized and contains two lone pairs of electrons. In OF2, the bond angle is reduced from OH2 as F being more electronegative, the bonded pairs will lie away from the central atom thus decreasing the repulsion. In OCl2, bonded pairs of electrons lie more close to O and increase the repulsion. ClO2 contains a three electron bond coupled with an electron pair bond. The shorter bond length and paramagnetic nature is due to three electron bond. The odd electron is not fully localized on the central atom. The larger bond angle 118° may be attributed to the existence of Cl—O π bonding which causes considerable delocalization of the odd electron thus decreasing the lone pair-lone pair repulsion.
Greater the pKa (—log Ka) value weaker the acid. Perchloric acid (HClO4) is the strongest acid (pKa = -8) followed by nitric acid (HNO3) (pKa = -1.4) and carbonic acid (H2CO3) (pKa = 7). Boric acid (H3BO3) is the weakest acid (pKa = 10)
The pKa values for the acids H2SO3 (pKa = 2), H3PO3 (pKa = 1.3) and HClO3 (pKa = -3) indicate that the decreasing order of acidity of these acids is HClO3 > H3PO3 > H2SO3.
Mg and F2 are prepared by electrolytic method.
ClO2 contains 7 + 12 i.e. 19 electrons (valence) which is an odd number.Hence it is paramagnetic in nature.
Conc. HCl reacts with KMnO4 solution most readily and gives Cl2 gas 2KMnO4 +16HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O
Calcium chloride is written as Ca (ClO2)2.
Electronegativity decreases from F to I and O is next to F in order of decreasing electronegativity.
NaClO4 on hydrolysis gives NaOH and strongest HClO4 acid as compared to other salts. Therefore pH of the solution will be lowest in this case.
RCOO‒ is not a pseudo halide (a group of two or more electronegative atoms).
As the oxidation number of the halogen increases, the acidity also increases.
Cu2+ gives iodometric titration as it gives I2 on reacting with KI.
Some of interhalogens are solids and are not volatile.
For the reaction as mentioned, all the three statements are correct.
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