We have to get the relationship between metallic character and atomic radius.
Metallic character increases with increase in atomic radius and decrease with decrease of atomic radius.
If electrons from outermost shell of an element can be removed easily, that atom can be considered to have more metallic character.
With increase in atomic radius, nuclear force of attraction towards outermost shell electron decreases which facilitates the release of electron.
With decrease in atomic radius, nuclear force of attraction towards outermost shell electrons increases, so electrons are hold tightly to nucleus. Hence, removal of electron from outermost shell becomes difficult making the atom less metallic in nature.
Electrons are only
about 0.054% as massive as neutrons and protons are only 99.86% as massive as
the neutrons. The mass of the Proton is 1.67 x 10^-27 kg and the mass of the electron
is 9.11 x 10^-31 kg. The mass of the electron is so much lighter than the mass
of the proton.
Answer:
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An atom of carbon has 4 electrons in its outermost shell, which means that
<span>its ionic charge is 4+ or 4-
</span>Si is in same group as carbon so its also 4+ or 4-
Germanium is 4+.
Sn is also 2+ or 4+
Pb is usually +2
Answer:
The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>
Explanation:
Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M
Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M
<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)
The equilibrium constant (
) for the given chemical reaction, is given by the equation:
![K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%20%7B%5BHI%5D%5E%7B2%7D%7D%7B%5BH_%7B2%7D%5D%5C%3A%20%5BI_%7B2%7D%5D%7D)
<u><em>At the original equilibrium state:</em></u>
<u><em>Therefore, at the new equilibrium state:</em></u>
![K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%20%7B%5BHI%5D%5E%7B2%7D%7D%7B%280.95%5C%3A%20M%29%20%5Ctimes%20%280.019%5C%3A%20M%29%7D)
![\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}](https://tex.z-dn.net/?f=%5CRightarrow%20K_%7Bc%7D%20%3D%20713.59%20%3D%20%5Cfrac%20%7B%5BHI%5D%5E%7B2%7D%7D%7B0.01805%7D)
![\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88](https://tex.z-dn.net/?f=%5CRightarrow%20%5BHI%5D%5E%7B2%7D%20%3D%20713.59%20%5Ctimes%200.01805%20%3D%2012.88)
![\Rightarrow [HI] = \sqrt {12.88} = 3.589 M](https://tex.z-dn.net/?f=%5CRightarrow%20%5BHI%5D%20%3D%20%5Csqrt%20%7B12.88%7D%20%3D%203.589%20M)
<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>
