Be + I2 → BeI2
It appears as colorless needle-like crystals and can be used in preparing high purity beryllium. Its molar mass is 262.81g/mol. It is soluble in diethyl ether, ethanol and slightly soluble in CS2. The reaction between BeI2 and water is very vigorous. A small trace of water in its surrounding can attack it and form hydroiodic acid. In this article, we will discuss beryllium Iodide Lewis dot structure, geometry, hybridization, polarity, bond angle, etc.
Lewis Dot Structure of BeI2
Lewis dot structure or electron dot structure is a representation of valence electrons on an individual atom as bond pairs or lone pairs, and the position of constituent atoms of a molecule with respect to each other. In a Lewis structure, Chemical symbols represent bonding atoms A dot represents a valence electron A pair of dots between bonding atoms represents a bond.
Steps to draw Lewis dot structure of BeI2
Step 1: Count the total number of valence electrons in BeI2: It’s easy to calculate the valence electrons of any element (up to atomic number 20) just by knowing its group number. BeI2 contains three atoms – 1 Beryllium atom and 2 Iodine atoms. Beryllium belongs to the 2nd group and Iodine belongs to the 17th group in the periodic table. Hence, Valence electrons in beryllium = 2 Valence electrons in iodine = 7 Total number of valence electrons available = 2 + 7 (2) = 16 Step 2: Choose the right central atom: The central atom needs to be with less electronegative value in the Lewis dot structure so that it can share more electrons with its surrounding atoms. Being an earth alkaline metal, beryllium is less electronegative than iodine in BeI2. Hence, put Be as the central atom and two iodine atoms as surrounding atoms. Step 3: Connect central atom and surrounding atoms via single bond: Draw a skeleton molecule of BeI2 and connect each surrounding atom (iodine) with the central atom (beryllium) via a single sigma bond. In this way, two single bonds are established. As a single bond denotes two electrons so, the number of valence electrons used in the above structure is 2×2 = 4 electrons. Number of valence electrons left: 16 – 4 = 12 valence electrons Step 4: Arrange the remaining valence electrons according to Octet Rule: The Octet rule states that the main group elements (s-block and p-block) in a molecule can transfer or share electrons till each atom acquires eight valence electrons to attain stability like noble gases.
After placing two single bonds, start arranging 12 valence electrons on the peripheral atoms (iodine) first to complete the octet. In case, some extra electrons remain, then place them on the central atom.
In this way, both iodine atoms acquire eight electrons in their valence (outermost) shells in the above structure. 6 valence electrons (represented as dots) and 2 electrons (represented as a single bond) are available to each iodine atom. However, the central atom (beryllium) has only 4 electrons in its valence shells. Like Hydrogen, Lithium, and Boron; the s-orbitals of the Beryllium atom do not require 8 valence electrons to complete the shell or to attain stability. So, all these elements contradict the octet rule. Step 5: Calculate the formal charge on each atom to check stability: The stability of the Lewis structure can be determined by calculating formal charges. A less formal charge on atoms indicates higher stability of the structure. In a neutral molecule, the net formal charge remains zero.
Easy ways to calculate formal charge:
• Formal charge = Valence electrons in a neutral atom – lone pair electrons – 1/2 bonded pair electrons • Formal charge = Group number – (number of unshared electrons + number of electron-pair bonds) In BeI2, both peripheral atoms (Iodine) have the same number of bonding electrons and lone pair electrons so, calculate the formal charge for any Iodine atom.
The formal charge on Iodine atom:
• Valence electrons of Iodine = 7 • Lone pair electrons on Iodine (represented as dots) = 6 • Bonded pair electrons around Iodine (represented as a single bond) = 2 • Formal charge on fluorine atoms: 7 – 6 – 2/2 = 0
The formal charge on Beryllium atom:
• Valence electrons of Beryllium = 2 • Lone pair electrons on Beryllium = 0 • Bonded pair electrons around Beryllium (represented as 2 single bonds) = 4 • Formal charge on Beryllium atom: 2 – 0 – 4/2 = 0 So, all the atoms in the BeI2 Lewis structure have zero formal charges. Hence, this is the most stable Lewis structure for BeI2.
BeI2 Geometry
VSEPR theory helps in predicting the geometry and shape of covalently bonded molecules like BeI2. Valence electron pair repulsion (VSEPR) theory states that the geometry of the molecule depends on the repulsion of electron pairs in the valence shells. Electrons always tend to align themselves at the maximum distance from each other to minimize the repulsions as well as to make the arrangement stable. The order of repulsive interaction between valence shell electrons is as follows: Lone pair (lp) – Lone pair (lp) ˃ Lone pair (lp) – Bond pair (bp) ˃ Bond pair (bp) – Bond pair (bp)
Steps for using VSEPR theory to determine the geometry of any molecule:
• Draw Lewis structure of the molecule • Count number of bonding electrons and lone pair electrons around the central atom Note: multiple bonds like double or triple bond count as only one bonding pair • Using table 1 or table 2, determine the molecular or electron geometry of the molecule Table 1 shows geometries of molecules without lone pair on the central atom:
Table 2 shows geometries of molecules with one or more lone pairs on the central atom In the Lewis structure of BeI2, Beryllium is the central atom and it has 2 bonded pairs (4 electrons) and no lone pair. So the molecular geometry and electron geometry for BeI2 remain the same. According to the above VSEPR table, its geometry is linear with a bond angle of 180⁰.
BeI2 Hybridization
Hybridization means the mixing of atomic orbitals of equivalent energies to form a new hybrid orbital of similar energy. Hybrid orbitals are oriented in such a way that the repulsion between electron pairs is minimum. Hybridization can be determined by calculating the number of included orbitals • ½ (number of monovalent atoms attached to central atom + number of valence electrons in a neutral central atom). • Number of sigma bonds of central atom + lone pairs on the central atom. As per the Lewis structure of BeI2, Beryllium forms two sigma single bonds with Iodine atoms with no lone pair. So, according to the above chart, BeI2 has sp hybridization. Or, it can be understood as the atomic number of Beryllium is 4 so, its electronic configuration is as follows: Be4 = 1s2 2s2 2pxo2py02pz0
Be has already completely filled s orbital in the ground state without any lone pair. So, it needs to excite its electron from 2s orbital to 2p orbital (2s and 2p have similar energy levels) to make a bond with Iodine. Its s and p orbital mixed up to form sp hybrid orbital to react with orbitals of Iodine
E.C of Iodine: 1s22s22p63s23p64s23d104p65s24d105p5 It has one unpaired electron in the p-orbital that reacts with the sp hybridized orbital of Be. Each unpaired electron in the p orbital of both Iodine atoms makes a bond with two unpaired electrons in sp hybridized orbitals of Be.
BeI2 Polarity
BeI2 is a nonpolar molecule due to its linear geometry. Symmetrical or unsymmetrical distribution of charges or separation of charges determines the polarity of the molecule. Polar molecules have non-zero dipole moments because of the un-symmetrical distribution of charges (H2O) or separation of charges on both ends ((Br-Cl). Whereas, non-polar molecules have the symmetrical distribution of charges because of their symmetrical structures leading to zero dipole moment. According to the Lewis structure of BeI2, the beryllium atom is attached to two iodine atoms. In BeI2, Both the bonds are arranged in the symmetrical order so the dipole moment generated on both sides (due to electronegativity difference between Be and I) cancels out each other making the molecule non-polar.
Conclusion
Beryllium Iodide is a covalent molecule in which beryllium forms two single bonds with Iodine atoms as peripheral atoms. Lewis structure of BeI2 has a total of 16 valence electrons in which 12 are lone pair electrons and 4 are bond pair electrons. Lewis structure of BeI2 does not follow the octet rule (Be has only 4 electrons in its valence shell) but it is stable due to zero formal charge. BeI2 has two bond pairs and zero lone pairs so based on VSEPR theory and hybridization, BeI2 has linear geometry and sp hybridization. Symmetrical geometry and uniform distribution of charges make BeI2 a non-polar molecule. Happy learning!!
