All categories

3 years ago

What atomic or hybrid orbitals make up the sigma bond between ge and h in germanium hydride, geh4 ?

2 answers:

ozzi3 years ago

8 0

Germane is the chemical compound with the formula GeH₄, and the germanium analogue of methane. It is the simplest germanium hydride and one of the most useful compounds of germanium.

In chemistry, sigma bonds (σ bonds) are the strongest type of covalent chemical bond. They are formed by head-on overlapping between atomic orbitals. Sigma bonding is most simply defined for diatomic molecules using the language and tools of symmetry groups.


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

ludmilkaskok [199]3 years ago

6 0

Answer:

Electron geometry: tetrahedral, sp3 hybridized.

Explanation:

Hello,

Since four atoms of hydrogen are surrounding the germanium in this case, a sp3 hybrization is making up the sigma bons for this molecular compound. The germanium atom can use its singly occupied p-type orbitals, to form two covalent bonds with two hydrogen atoms, and use the s and the other p to form the other two covalent bonds with the remaining hydrogens.

Best regards.

You might be interested in

PLEASE HELP MEEEE!!!! 10 POINTS + BRAINLYEST!!!!!!

SIZIF [17.4K]

Answer:

God's power is not reflected here.

5 0

2 years ago

How many formula units make up 24.2 g of magnesium chloride (MgCl2)? Help!!

NNADVOKAT [17]

Answer:

Approximately $1.53 \times 10^{23}$ formula units ( $0.254\; \rm mol$ ).

Explanation:

Refer to a modern periodic table for the relative atomic mass of magnesium ( $\rm Mg$ ) and chlorine ( $\rm Cl$ ):

$\rm Mg$ : $24.305$ .
$\rm Cl$ : $35.45$ .

In other words, the mass of $1\; \rm mol$ of $\rm Mg$ atoms would be (approximately) $24.305\; \rm g$ .

Likewise, the mass of $1\; \rm mol$ of $\rm Cl$ atoms would be approximately $35.45\; \rm g$ .

One formula unit of the ionic compound $\rm MgCl_{2}$ includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of $1\; \rm mol$ of the formula units of this compound.

The formula $\rm MgCl_{2}$ includes one $\rm Mg$ atom and two $\rm Cl$ atoms.

Hence, every formula unit of $\rm MgCl_{2} \!$ would include the same number of atoms: one $\rm Mg\!$ atom and two $\rm Cl\!$ atoms. There would be $1\; \rm mol$ of $\rm Mg$ atoms and $2\; \rm mol$ of $\rm Cl$ atoms in $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units.

Thus, the mass of $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units would be equal to the mass of $1\; \rm mol$ of $\rm Mg$ atoms plus the mass of $2\; \rm mol$ of $\rm Cl$ atoms. (The mass of $1\; \rm mol\!\!$ of each atom could be found from the relative atomic mass of each element.)

$\begin{aligned}& M({\rm MgCl_{2}}) \\ =\; & 24.305\; {\rm g \cdot mol^{-1}} + 2\times {\rm 35.45 \; \rm g \cdot mol^{-1}} \\ =\; & 95.205\; \rm g \cdot mol^{-1}\end{aligned}$ .

In other words, the formula mass of $\rm MgCl_{2}$ is $95.205\; \rm g \cdot mol^{-1}$ .

Therefore, the number of formula units in $m = 24.2\; \rm g$ of $\rm MgCl_{2}$ would be:

$\begin{aligned}n &= \frac{m({\rm MgCl_{2}})}{M({\rm MgCl_{2}})} \\ &= \frac{24.2\; \rm g}{95.205\; \rm g\cdot mol^{-1}} \\ & \approx 0.254\; \rm mol\end{aligned}$ .

Multiple $n$ by Avogadro's Number $N_{A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}$ to estimate the number of formula units in $0.254\; \rm mol$ :

$\begin{aligned}N &= n \cdot N_{A} \\ &\approx 0.254\; \rm mol \times 6.022 \times 10^{23}\; \rm mol^{-1} \\ &\approx 1.53\times 10^{23}\end{aligned}$ .

6 0

3 years ago

Choose the answer that best completes the following statement: When an aluminum atom reacts so as to attain a noble gas electron

boyakko [2]

The options

Select one:

a. a 3- ion forms.

b. the noble gas configuration of argon is achieved.

c. the result is a configuration of 1s2 2s2 2p6.

d. the atom gains five electrons.

Answer:

c. the result is a configuration of 1s2 2s2 2p6.

Explanation:

Aluminium atom has atomic number of 13 , hence the number of electron is 13 for a neutral atom of aluminium. When aluminium atom reacts with other elements it usually gives out three electron to attain the octet configuration.

The cation representation of aluminium is Al3+ because it has loss three electron to attain the octet rule. Aluminium will be left with 10 electrons after losing 3 of it electrons. The electronic configuration will be represented as follows after losing three electrons;

1S² 2S² 2P∧6 .

At this stage the octet rule has been achieved as it will be represented as

2 8. The first energy shell now contains two electron and the second energy shell contains 8 electrons.

The configuration of Neon has been formed in the process.

4 0

3 years ago

How do kinetic tiles work?

raketka [301]

Answer:

It works because of the significant principle of piezoelectricity

Explanation:

7 0

4 years ago

Does H₂O have a subscript?

erma4kov [3.2K]

2 hydrogen atoms attached to an oxygen atom.

7 0

3 years ago

Read 2 more answers