Answer:
In the Lewis structure of P4 there are 6 bonding pairs and 4 lone pairs of electrons.
Explanation:
The structure of tetrahedral molecule of P4 is provided below.
Each phosphorus atom has 5 valence electrons out of which 3 electrons involve in bonding and the rest 2 electrons exist as a lone pair that does not involve in bonding.Hence each phosphorus atom has one lone pair.In P4 molecule there are phosphorus atoms and hence 4 lone pairs in total.
As you can see in the figure, each phosphorus atom is bonded to the other three atoms.A bond is formed when two atoms share one electron each and the pair is called bonding pair.
From the figure we can see that there are 6 bonds in total.Each bond consist of one bonding pair of electrons and hence in total there are 6 bonding pairs of electrons.
Hence in a P4 molecule there are six bonding pairs and 4 lone pairs of electrons.
Answer:
Change in internal energy (ΔU) = -9 KJ
Explanation:
Given:
q = –8 kJ [Heat removed]
w = –1 kJ [Work done]
Find:
Change in internal energy (ΔU)
Computation:
Change in internal energy (ΔU) = q + w
Change in internal energy (ΔU) = -8 KJ + (-1 KJ)
Change in internal energy (ΔU) = -8 KJ - 1 KJ
Change in internal energy (ΔU) = -9 KJ
The elements in the periodice table are not listed in alphabetical order, because the arragement in rows (periods) and columns (groups or familes), in increasing order of atomic number (number of protons of the atoms) permits to explain similarities among the elements, trend in some properties, and even predict properties of unknown elements.
For example, the elements of the first group (family), called alkaline metals, all have 1 valence electron, have similar physical properties (ductibility, malleability, luster, thermal and electricity conductivity), react in similar way with water, show a trend in the atomic radii and in the ionization energy.
You can tell similar stories for other groups like, alkalyne earth metals, halogens and noble gases.
You can also tell trends in electroneativities, and atomic radii, for a row of elements, as per the order they are in the row.
So, the current array resulted very helpul for chemists to explain and predict the behavior and properties of the elements.
Answer:
p3=0.36atm (partial pressure of NOCl)
Explanation:
2 NO(g) + Cl2(g) ⇌ 2 NOCl(g) Kp = 51
lets assume the partial pressure of NO,Cl2 , and NOCl at eequilibrium are P1 , P2,and P3 respectively
![Kp=\frac{[NOCl]^{2} }{[NO]^{2} [Cl_2] }](https://tex.z-dn.net/?f=Kp%3D%5Cfrac%7B%5BNOCl%5D%5E%7B2%7D%20%7D%7B%5BNO%5D%5E%7B2%7D%20%5BCl_2%5D%20%7D)
![Kp=\frac{[p3]^{2} }{[p1]^{2} [p2] }](https://tex.z-dn.net/?f=Kp%3D%5Cfrac%7B%5Bp3%5D%5E%7B2%7D%20%7D%7B%5Bp1%5D%5E%7B2%7D%20%5Bp2%5D%20%7D)
p1=0.125atm;
p2=0.165atm;
p3=?
Kp=51;
On solving;
p3=0.36atm (partial pressure of NOCl)
Magnetic moment (spin only) of octahedral complex having CFSE=−0.8Δo and surrounded by weak field ligands can be : Q
To answer this, the Crystal Field Stabilization Energy has to be calculated for a (d3 metal in both configurations. The geometry with the greater stabilization will be the preferred geometry. So for tetrahedral d3, the Crystal Field Stabilization Energy is: CFSE = -0.8 x 4/9 Δo = -0.355 Δo.
[Co(CN)64-] is also an octahedral d7 complex but it contains CN-, a strong field ligand. Its orbital occupancy is (t2g)6(eg)1 and it therefore has one unpaired electron. In this case the CFSE is −(6)(25)ΔO+(1)(35)ΔO+P=−95ΔO+P.
The crystal field stabilization energy (CFSE) (in kJ/mol) for complex, [Ti(H2O)6]3+. According to CFT, the first absorption maximum is obtained at 20,3000cm−1 for the transition.
To learn more about crystal field stabilization energy visit:brainly.com/question/29389010
#SPJ4
