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Sia ? 6 years, 6 months ago
have large charge/radius ratio;
are hard and have high densities;
have high melting and boiling points;
form compounds which are often paramagnetic;
show variable oxidation states;
form coloured ions and compounds;
form compounds with profound catalytic activity;
Sia ? 6 years, 6 months ago
Elements in which the last electron enters the d-orbitals, are called d-block elements. These elements have general outer electronic configuration (n-1)d1-10 ns0-2. Zn, Cd and Hg which have general electronic configuration, (n-1) d10ns2 do not show most of the properties of transition elements. d-orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Thus, on the basis of properties, all transition elements are d-block elements but on the basis of electronic configuration, all d- block elements are not transition elements.
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Due to greater electronegativity of O than S, H2O undergoes extensive intermolecular H-bonding. As a result, H2O is a liquid at room temperature. In contrast, H2S does not undergo H-bonding. It exists as discrete molecules which are held together by weak Van der Waals forces of attraction. To break these forces of attraction, only a small amount of energy is required. Therefore, H2S is a gas at room temperature.
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Sia ? 6 years, 6 months ago
Elements in which the last electron enters the d-orbitals, are called d-block elements. These elements have general outer electronic configuration (n-1)d1-10 ns0-2. Zn, Cd and Hg which have general electronic configuration, (n-1) d10ns2 do not show most of the properties of transition elements. d-orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Thus, on the basis of properties, all transition elements are d-block elements but on the basis of electronic configuration, all d- block elements are not transition elements.
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Sia ? 6 years, 6 months ago
Given Percentage of C= 40.687% , Percentage of H = 5.085%.
Therefore, Percentage of O = 100 - (40.687+5.085) = 54.228%.
<th scope="col">Element</th> <th scope="col">Symbol</th> <th scope="col">Percentage of element</th> <th scope="col">Moles of the elementStep I: To calculate the empirical formula of the compound.
Since, ration of C : H : O = 2 :3 :2.
{tex}\therefore{/tex} An empirical formula is C2H3O2.
Step II: The empirical formula of the compound = C2H3O2.
{tex}\therefore{/tex} Empirical formula mass = 2 {tex}\times{/tex} C +3 {tex}\times{/tex} H + 2{tex}\times{/tex} O = {tex}( 2 \times 12 ) + ( 3 \times 1 ) + ( 2 \times 16 ) = 59{/tex}
Step III: To calculate the molecular mass of the salt
The vapour density of the compound = 59 (Given)
Using the relation between vapour density and molecular mass.
Therefore, Molecular mass of compound = 2 {tex}\times{/tex} vapour density of compound = 2 {tex}\times{/tex} 59 = 118
Step IV: The value of n = {tex}\frac { \text { molecular mass } } { \text { empirical formula mass } } = \frac { 118 } { 59 } = 2{/tex}
Step V: Calculation of the molecular formula of the salt,
Molecular formula = n {tex}\times{/tex} empirical formula = {tex}2 \times \mathrm { C } _ { 2 } \mathrm { H } _ { 3 } \mathrm { O } _ { 2 } = \mathrm { C } _ { 4 } \mathrm { H } _ { 6 } \mathrm { O } _ { 4 }{/tex}
Thus, the molecular formula is C4H6O4.
3Thank You