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3 years ago

What is the electron configuration of the element in period 2 that has 6 valence electrons?

Chemistry

1 answer:

kompoz [17]3 years ago

8 0

Answer:

The order of a element in period 2 that has 6 valence electrons is:

1s²2s²2p²

Explanation:

The Electronic Configuration of the elements is the arrangement of all electrons of an element in energy levels and sub-levels (orbitals).

There are 7 energy levels, numbered from 1 to 7, and in which electrons are distributed, logically in order according to their energy level. Electrons with less energy will be spinning at level 1. Each level is divided into sub-levels. These sub-levels into which each level is divided can be up to 4. These 4 sub-levels are called: s, p, d, f. In the sub-level s there can only be a maximum of 2 electrons, in p there can be a maximum of 6 electrons, in the sub-level d 10 electrons and finally in the sub-level f there can be a maximum of 14 electrons.

In level 1 there is only one sub-level, which will be the s. In level 2 there are 2 sub-levels, the s and the p. At level 3 there are 3 sub-levels s, p and d. And at level 4 there are 4 sub-levels, the s, the p, the d and the f.

Aufbau's principle is a principle of atomic physics, which explains how electrons are accommodated in their orbits around the nucleus of the atom. This principle states that the filling of these orbitals occurs in increasing order of energy, that is, from the orbitals of lower energy to those of higher energy.

The simplest of the sub-levels is 1s. If you fill the 1s you can start filling the 2s. After the 2s sub-level is satisfied, the 2p can begin to fill and so on.

So, the order of a element in period 2 that has 6 valence electrons is:

1s²2s²2p²

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3 years ago

Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp

Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

$\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}$

For calculating edge length,

$a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}$

For calculating $M_{ave}$ , we use the formula

$M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}$

Similarly for calculating $(\rho)_{ave}$ , we use the formula

$\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}$

From the periodic table, masses of the two elements can be written

$M_{Fe}= 55.85g/mol$

$M_{V}=50.941g/mol$

Specific density of both the elements are

$(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3$

Putting $M_{ave}$ and $\rho_{ave}$ formula's in edge length formula, we get

$a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}$

$a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}$