K12? Nice! Eh. Idk, I will comment the answer when I get it
Explanation:
Octahedral complexes will be favoured over tetrahedral ones because:
It is more favourable to form six bonds rather than four
The crystal field stabilisation energy is usually greater for octahedral than tetrahedral complexes.
The transition metals Co, Rh, Lr are in group 9 of d block and they have 3d, 4d, and 5d orbitals respectively
The molecule BH3 is trigonal planar, with B in the center and H in the three vertices. Ther are no free electrons. All the valence electrons are paired in and forming bonds.
There are four kind of intermolecular attractions: ionic, hydrogen bonds, polar and dispersion forces.
B and H have very similar electronegativities, Boron's electronegativity is 2.0 and Hydrogen's electronegativity is 2.0.
The basis of ionic compounds are ions and the basis of polar compounds are dipoles.
The very similar electronegativities means that B and H will not form either ions or dipoles. So, that discards the possibility of finding ionic or polar interactions.
Regarding, hydrogen bonds, that only happens when hydrogen bonds to O, N or F atoms. This is not the case, so you are sure that there are not hydrogen bonds.
When this is the case, the only intermolecular force is dispersion interaction, which present in all molecules.
Then, the answer is dispersion interaction.
Answer:
5.96 g/cm^3
Explanation:
Corner atom = 1/8
Atoms in center = 1
Atoms in face of the cube= 1/2
Molar mass of V = 50.94 g/mol <em>(from period table)</em>
1 mole = 6.02x10^23
<em>In BCC unit cell:</em>
(8 x 1/8)+ 1=2 per 1 unit cell
<em>Mass: </em>2(50.94g)/6.02x10^23 = 1.69x10^-22 g/unit cell
305pm=(305x10^-12m÷10^-2m) x (1mL÷1cm^3)
= 2.837 x 10^-23 mL
<em>1pm=10^-12m</em>
<em>1cm=10^-2m</em>
<em>1mL=1cm^3</em>
<em></em>
density=mass/volume
density of V = 1.69x10^-22g÷2.837x10^-23mL
=5.957g/mL
=5.96g/cm^3
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