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Nezavi [6.7K]
3 years ago
14

All p block elements form ions. True or False

Chemistry
1 answer:
FinnZ [79.3K]3 years ago
4 0

Answer:

No this statement is false.

Explanation:

when an atom lose or gain the electron ions are formed. If the atom lose an electron the positive ions are formed called cations while anion is formed by the gaining of electron by an atom The elements having less electrons in valance shell usually lose their electrons while the elements like halogens having 7 valance electrons gain an electron to complete the octet. while p block elements noble gases are inert, their outer most valance shell is complete so they do not form ions.

P block elements are non-metals, metals and metalloids. These are thirty five elements. The P-block elements are present on right side of periodic table. There valance electrons are present in P orbital. The p-block metals are shiny and good conductor of heat and electricity. These metal lose the electron which is accept by non metals and form ionic bond. They have high melting points.

Metalloids includes boron, silicon, germanium, arsenic, antimony and tellurium. Metalloids contain both the properties of metals and non metals, Some metalloids are toxic like arsenic.

Most of p-block elements are non metals. They are bad conductor of heat and electricity and have low boiling points. The non metals mostly accept the electron from the metals and usually from ionic bond like in case of chlorine. It form the ionic compound with sodium.

The sodium chloride which is an ionic compound, formed by the complete transfer of electron from sodium to chlorine atom and form ionic bond. In this ionic compound sodium carry positive charge and chlorine carry negative charge there is attraction between these oppositely charged atoms.

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VikaD [51]

Answer : The value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

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Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)

The half-cell reactions are:

Oxidation reaction (anode) : Pb(s)\rightarrow Pb^{2+}(aq)+2e^-

Reduction reaction (cathode) : 2Ag^+(aq)+2e^-\rightarrow 2Ag(g)

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

\Delta G^o=-2\times 96500\times 0.93

\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol

Now we have to calculate the value of 'K'.

\Delta G^o=-RT\ln K

where,

\Delta G_^o =  standard Gibbs free energy  = -180 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K

K=3.6\times 10^{31}

Therefore, the value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

5 0
3 years ago
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